ML Aggarwal Solutions for Chapter 3 Expansions Class 9 Maths ICSE
Exercise 3.1
By using standard formulae, expand the following (1 to 9):
1. (i) (2x + 7y)2
(ii) (1/2 x + 2/3 y)2
Solution
(i) (2x + 7y)2
It can be written as
= (2x)2 + (2×2x ×7y) + (7y)2
So we get,
= 4x2 + 28xy + 49y2
(ii) (1/2 x + 2/3 y)2
It can be written as
= (1/2 x)2 + (2×½x) + 2/3 y + (2/3 y)2
So we get,
= ¼ x2 + 2/3 xy + 4/9 y2
2. (i) (3x + 1/2x)2
(ii) (3x2y + 5z)2
Solution
(i) (3x + 1/2x)2
It can be written as
= (3x)2 + (2×3x×1/2 x) + (1/2x)2
So we get
= 9x2 + 3 + 1/4x2
= 9x2 + 1/4x2 + 3
(ii) (3x2y + 5z)2
It can be written as
= (3x2y)2 + (2×3x2y×5z) + (5z)2
So we get
= 9x4y2 + 30x2yz + 25z2
3. (i) (3x – 1/2x)2
(ii) (1/2 x – 3/2 y)2
Solution
(i) (3x – 1/2x)2
It can be written as
= (3x)2 – (2×3x×1/2 x) + (1/2x)2
So we get
= 9x2 – 3 + 1/4x2
= 9x2 + 1/4x2 – 3
(ii) (1/2 x – 3/2 y)2
It can be written as
= (1/2 x)2 + (3/2 y)2 – (2×½ x×3/2 y)
So we get,
= ¼ x2 + 9/4 y2 – 3/2 xy
= ¼ x2 – 3/2 xy + 9/4 y2
4. (i) (x + 3) (x + 5)
(ii) (x + 3) (x – 5)
(iii) (x – 7) (x + 9)
(iv) (x – 2y) (x – 3y)
Solution
(i) (x + 3) (x + 5)
By further calculation
= x2 + (3 + 5) x + 3 × 5
So we get
= x2 + 8x + 15
(ii) (x + 3) (x – 5)
By further calculation
= x2 + (3 – 5)x – 3 × 5
So we get
= x2 – 2x – 15
(iii) (x – 7) (x + 9)
By further calculation
= x2 – (7 – 9)x – 7 × 9
So we get
= x2 + 2x – 63
(iv) (x – 2y) (x – 3y)
By further calculation
= x2 – (2y + 3y)x + 2y × 3y
So we get
= x2 – 5xy + 6y2
5. (i) (x – 2y – z)2
(ii) (2x – 3y + 4z)2
Solution
(i) (x – 2y – z)2
It can be written as
= [x + (-2y) + (-z)]2
By further calculation
= (x)2 + (-2y)2 + (-z)2 + {2×x×(-2y)} + {2×(-2y)×(-z)} + {2×(-z)×x}
So we get
= x2 + 4y2 + z2 – 4xy + 4yz – 2zx
(ii) (2x – 3y + 4z)2
It can be written as
= [2x + (-3y) + 4z]2
By further calculation
= (2x)2 + (-3y)2 + (4z)2 + {2×2x×(-3y)} + {2×(-3y)×4z} + {2×4z×2x}
So we get
= 4x2 + 9y2 + 16z2 -12xy – 24yz + 16zx
6. (i) (2x + 3/x – 1)2
(ii) (2/3 x – 3/2x – 1)2
Solution
(i) (2x + 3/x – 1)2
It can be written as
= [2x + 3/x + (-1)]2
By further calculation
= (2x)2 + (3/x)2 + (-1)2 + {2×2x× 3/x} + {2× 3/x ×(-1)} + {2×(-1)×2x}
So we get
= 4x2 + 9/x2 + 1 + 12 – 6/x – 4x
= 4x2 + 9/x2 + 13 – 6/x – 4x
(ii) (2/3 x – 3/2x – 1)2
It can be written as
= [2/3 x – 3/2x – 1]2
By further calculation
= (2/3 x)2 + (-3/2x)2 + (-1)2 + {2 × 2/3 x ×(-3/2x)} + {2×(-3/2x)×(-1)} + {2×(-1)×(2/3 x)}
So we get
= 4/9 x2 + 9/4x2 + 1 – 2 + 3/x – 4/3 x
= 4/9 x2 + 9/4x2 – 1 – 4/3 x + 3/x
7. (i) (x + 2)3
(ii) (2a + b)3
Solution
(i) (x + 2)3
It can be written as
= x3 + 23 + {3×x×2 (x + 2)}
By further calculation
= x3 + 8 + 6x(x + 2)
So we get
= x3 + 8 + 6x2 + 12x
= x3 + 6x2 + 12x + 8
(ii) (2a + b)3
It can be written as
= (2a)3 + b3 + {3×2a×b(2a + b)}
By further calculation
= 8a3 + b3 + 6ab (2a + b)
So we get
= 8a3 + b3 + 12a2b + 6ab2
8. (i) (3x + 1/x)3
(ii) (2x – 1)3
Solution
(i) (3x + 1/x)3
It can be written as
= (3x)3 + (1/x)3 + {3×3x× 1/x(3x + 1/x)}
By further calculation
= 27x3 + 1/x3 + 9 (3x + 1/x)
So we get
= 27x3 + 1/x3 + 27x + 9/x
(ii) (2x – 1)3
It can be written as
= (2x)3 – 13 – {3×2x× 1(2x–1)}
By further calculation
= 8x3 – 1 – 6x(2x – 1)
So we get
= 8x3 – 1 – 12x2 + 6x
= 8x3 – 12x2 + 6x – 1
9. (i) (5x – 3y)3
(ii) (2x – 1/3y)3
Solution
(i) (5x – 3y)3
It can be written as
= (5x)3 – (3y)3 – {3×5x× 3y(5x–3y)}
By further calculation
= 125x3 – 27y3 – 45xy (5x – 3y)
So we get
= 125x3 – 27y3 – 225x2y + 135xy2
(ii) (2x – 1/3y)3
It can be written as
= (2x)3 – (1/3y)3 – {3×2x× 1/3y (2x–1/3y)}
By further calculation
= 8x3 – 1/27y3 – 2x/y (2x– 1/3y)
So we get
= 8x3 – 1/27y3 – 4x2/y + 2x/3y2
Simplify the following (10 to 19):
10. (i) (a + b)2 + (a – b)2
(ii) (a + b)2 – (a – b)2
Solution
(i) (a + b)2 + (a – b)2
It can be written as
= (a2 + b2 + 2ab) + (a2 +b2 – 2ab)
By further calculation
= a2 + b2 + 2ab + a2 + b2 – 2ab
So we get
= 2a2 + 2b2
Taking 2 as common
= 2 (a2 + b2)
(ii) (a + b)2 – (a – b)2
It can be written as
= (a2 + b2 + 2ab) – (a2 + b2 – 2ab)
By further calculation
= a2 + b2 + 2ab – a2 – b2 + 2ab
So we get
= 4ab
11. (i) (a + 1/a)2 + (a – 1/a)2
(ii) (a + 1/a)2 – (a – 1/a)2
Solution
(i) (a + 1/a)2 + (a – 1/a)2
It can be written as
= [a2 + (1/a)2 + 2×a× 1/a] + [a2 + (1/a)2 – 2×a× 1/a]
By further calculation
= [a2 + 1/a2 + 2] + [a2 + 1/a2 – 2]
So we get
= a2 + 1/a2 + 2 + a2 + 1/a2 – 2
= 2a2 + 2/a2
Taking 2 as common
= 2 (a2 + 1/a2)
(ii) (a + 1/a)2 – (a – 1/a)2
It can be written as
= [a2 + (1/a)2 + 2×a× 1/a] – [a2 + (1/a)2 – 2×a× 1/a]
By further calculation
= [a2 + 1/a2 + 2] – [a2 + 1/a2 – 2]
So we get
= a2 + 1/a2 + 2 – a2 – 1/a2 + 2
= 4
12. (i) (3x – 1)2 – (3x – 2) (3x + 1)
(ii) (4x + 3y)2 – (4x – 3y)2 – 48xy
Solution
(i) (3x – 1)2 – (3x – 2) (3x + 1)
It can be written as
= [(3x)2 + 12 – 2×3x×1] – [(3x)2 – (2–1)(3x) – 2×1]
By further calculation
= [9x2 + 1 – 6x] – [9x2 – 3x – 2]
So we get
= 9x2 + 1 – 6x – 9x2 + 3x + 2
= -3x + 3
= 3 – 3x
(ii) (4x + 3y)2 – (4x – 3y)2 – 48xy
It can be written as
= [(4x)2 + (3y)2 + 2×4x×4y] – [(4x)2 + (3y)2 – 2×4x×3y] – 48xy
By further calculation
= [16x2 + 9y2 + 24xy] – [16x2 + 9y2 – 24xy] – 48xy
So we get
= 16x2 + 9y2 + 24xy – 16x2 – 9y2 + 24xy – 48xy
= 0
13. (i) (7p + 9q) (7p – 9q)
(ii) (2x – 3/x) (2x + 3/x)
Solution
(i) (7p + 9q) (7p – 9q)
It can be written as
= (7p)2 – (9q)2
= 49p2 – 81q2
(ii) (2x – 3/x) (2x + 3/x)
It can be written as
= (2x)2 – (3/x)2
= 4x2 – 9/x2
14. (i) (2x – y + 3) (2x – y – 3)
(ii) (3x + y – 5) (3x – y – 5)
Solution
(i) (2x – y + 3) (2x – y – 3)
It can be written as
= [(2x – y) + 3] [(2x – y) – 3]
= (2x – y)2 – 32
By further calculation
= (2x)2 +y2 – 2 × 2x × y – 9
So we get
= 4x2 + y2 – 4xy – 9
(ii) (3x + y – 5) (3x – y – 5)
It can be written as
= [(3x – 5) + y] [(3x – 5) – y]
= (3x – 52) – y2
By further calculation
= (3x)2 + 52 – 2 × 3x × 5 – y2
So we get
= 9x2 + 25 – 30x – y2
= 9x2 – y2 – 30x + 25
15. (i) (x + 2/x – 3) (x – 2/x – 3)
(ii) (5 – 2x) (5 + 2x) (25 + 4x2)
Solution
(i) (x + 2/x – 3) (x – 2/x – 3)
It can be written as
= [(x – 3) + (2/x)] [(x – 3) – (2/x)]
= (x – 3)2 – (2/x)2
Expanding using formula
= x2 + 9 – 2 × x × 3 – 4/x2
By further calculation
= x2 + 9 – 6x – 4/x2
So we get
= x2 – 4/x2 – 6x + 9
(ii) (5 – 2x) (5 + 2x) (25 + 4x2)
It can be written as
= [52 – (2x)2] (25 + 4x2)
By further calculation
= (25 – 4x2) (25 + 4x2)
So we get
= 252 – (4x2)2
= 625 – 16x4
16. (i) (x + 2y + 3) (x + 2y + 7)
(ii) (2x + y + 5) (2x + y – 9)
(iii) (x – 2y – 5) (x – 2y + 3)
(iv) (3x – 4y – 2) (3x – 4y – 6)
Solution
(i) (x + 2y + 3) (x + 2y + 7)
Consider x + 2y = a
(a + 3) (a + 7) = a2 + (3 + 7) a + 3×7
By further calculation
= a2 + 10a + 21
Substituting the value of a
= (x + 2y)2 + 10 (x + 2y) + 21
By expanding using formula
= x2 + 4y2 + 2×x×2y + 10x + 20y + 21
So we get
= x2 + 4y2 + 4xy + 10x + 20y + 21
(ii) (2x + y + 5) (2x + y – 9)
Consider 2x + y = a
(a + 5) (a – 9) = a2 + (5 – 9) a + 5×(-9)
By further calculation
= a2 – 4a – 45
Substituting the value of a
= (2x + y)2 – 4 (2x + y) – 45
By expanding using formula
= 4x2 + y2 + 2×2x×y – 8x – 4y – 45
So we get
= 4x2 + y2 + 4xy – 8x – 4y – 45
(iii) (x – 2y – 5) (x – 2y + 3)
Consider x – 2y = a
(a – 5) (a + 3) = a2 + (- 5 + 3) a + (-5) (3)
By further calculation
= a2 – 2a – 15
Substituting the value of a
= (x – 2y)2 – 2 (x – 2y) – 15
By expanding using formula
= x2 + 4y2 – 2×x×2y – 2x + 4y – 15
So we get
= x2 + 4y2 – 4xy – 2x + 4y – 15
(iv) (3x – 4y – 2) (3x – 4y – 6)
Consider 3x – 4y = a
(a – 2) (a – 6) = a2 (-2 – 6)a + (-2)(-6)
By further calculation
= a2 – 8a + 12
Substituting the value of a
= (3x – 4y)2 – 8(3x – 4y) + 12
Expanding using formula
= 9x2 + 16y2 – (2×3x×4y) – 24x + 32y + 12
So we get
= 9x2 + 16y2 – 24xy – 24x + 32y + 12
17. (i) (2p + 3q) (4p2 – 6pq + 9q2)
(ii) (x + 1/x) (x2 – 1 + 1/x2)
Solution
(i) (2p + 3q) (4p2 – 6pq + 9q2)
It can be written as
= (2p + 3q) [(2p)2 – (2p×3q) + (3q)2]
By further simplification
= (2p)3 + (3q)3
= 8p3 + 27q3
(ii) (x + 1/x) (x2 – 1 + 1/x2)
It can be written as
= (x + 1/x) [x2 – x × 1/x + (1/x)2]
By further simplification
= x3 + (1/x)3
= x3 + 1/x3
18. (i) (3p – 4q) (9p2 + 12pq + 16q2)
(ii) (x – 3/x) (x2 + 3 + 9/x2)
Solution
(i) (3p – 4q) (9p2 + 12pq + 16q2)
It can be written as
= (3p – 4q) [(3p)2 + (3p×4q) + (4q)2]
By further simplification
= (3p)3 – (4q)3
= 27p3 – 64q3
(ii) (x – 3/x) (x2 + 3 + 9/x2)
It can be written as
= (x – 3/x) [x2 + (x× 3/x) + (3/x)2]
By further simplification
= x3 – (3/x)3
= x3 – 27/x3
19. (2x + 3y + 4z) (4x2 + 9y2 + 16z2 – 6xy – 12yz – 8zx).
Solution
(2x + 3y + 4z) (4x2 + 9y2 + 16z2 – 6xy – 12yz – 8zx)
It can be written as
= (2x + 3y + 4z) ((2x)2 + (3y)2 + (4z)2 – (2x×3y) – (3y×4z) – (4z×2x)
By further calculation
= (2x)3 + (3y)3 + (4z)3 – (3×2x×3y×4z)
So we get
= 8x3 + 27y3 + 64z3 – 72xyz
20. Find the product of the following:
(i) (x + 1) (x + 2) (x + 3)
(ii) (x – 2) (x – 3) (x + 4)
Solution
(i) (x + 1) (x + 2) (x + 3)
It can be written as
= x3 + (1 + 2 + 3)x2 + (1×2) + (2×3) + (3×1) x + (1×2×3)
By further calculation
= x3 + 6x2 + (2+6+3)x + 6
So we get
= x3 + 6x2 + 11x + 6
(ii) (x – 2) (x – 3) (x + 4)
It can be written as
= x3 + (- 2 – 3 + 4) x2 + [(-2)×(-3) + (-3)×4 + 4×(-2)]x + (-2) (-3) (4)
By further calculation
= x3 – x2 + (6 – 12 – 8)x + 24
= x3 – x2 – 14x + 24
21. Find the coefficient of x2 and x in the product of (x – 3) (x + 7) (x – 4).
Solution
It is given that
(x – 3) (x + 7) (x – 4)
By further calculation
= x3 + (-3+7–4)x2 + [(-3) (7) + 7×(-4) + (-4)(-3) + (-3)(7)(-4)]
It can be written as
= x3 + 0x2 + (-21–28+12)x + 84
So we get
= x3 + 0x2 – 37x + 84
Hence, coefficient of x2 is zero and coefficient of x is – 3.
22. If a2 + 4a + x = (a + 2)2, find the value of x.
Solution
It is given that
a2 + 4a + x = (a + 2)2
By expanding using formula
a2 + 4a + x = a2 + 22 + (2×a×2)
By further calculation
a2 + 4a + x = a2 + 4 + 4a
So we get,
x = a2 + 4 + 4a – a2 – 4a
⇒ x = 4
23. Use (a + b)2 = a2 + 2ab + b2 to evaluate the following:
(i) (101)2
(ii) (1003)2
(iii) (10.2)2
Solution
(i) (101)2
It can be written as
= (100 + 1)2
Expanding using formula
= 1002 + 12 + (2×100×1)
By further calculation
= 10000 + 1 + 200
= 10201
(ii) (1003)2
It can be written as
= (1000 + 3)2
Expanding using formula
= 10002 + 32 + (2×1000×3)
By further calculation
= 1000000 + 9 + 6000
= 1006009
(iii) (10.2)2
It can be written as
= (10 + 0.2)2
Expanding using formula
= 102 + 0.22 + (2×10×0.2)
By further calculation
= 100 + 0.04 + 4
= 104.04
24. Use (a – b)2 = a2 – 2ab – b2 to evaluate the following:
(i) (99)2
(ii) (997)2
(iii) (9.8)2
Solution
(i) (99)2
It can be written as
= (100 – 1)2
Expanding using formula
= 1002 – (2×100×1) + 12
By further calculation
= 10000 – 200 + 1
= 9801
(ii) (997)2
It can be written as
= (1000 – 3)2
Expanding using formula
= 10002 – (2×1000×3) + 32
By further calculation
= 1000000 – 6000 + 9
= 994009
(iii) (9.8)2
It can be written as
= (10 – 0.2)2
Expanding using formula
= 102 – (2×10×0.2) + 0.22
By further calculation
= 100 – 4 + 0.04
= 96.04
25. By using suitable identities, evaluate the following:
(i) (103)3
(ii) (99)3
(iii) (10.1)3
Solution
(i) (103)3
It can be written as
= (100 + 3)3
Expanding using formula
= 1003 + 33 + {3×100×3(100 + 3)}
By further calculation
= 1000000 + 27 + (900×103)
So we get
= 1000000 + 27 + 92700
= 1092727
(ii) (99)3
It can be written as
= (100 – 1)3
Expanding using formula
= 1003 – 13 – {3×100×1(100–1)}
By further calculation
= 1000000 – 1 – (300×99)
So we get
= 1000000 – 1 – 29700
= 1000000 – 29701
= 970299
(iii) (10.1)3
It can be written as
= (10 + 0.1)3
Expanding using formula
= 103 + 0.13 + {3×10×0.1(10+0.1)}
By further calculation
= 1000 + 0.001 + (3×10.1)
So we get
= 1000 + 0.001 + 30.3
= 1030.301
26. If 2a – b + c = 0, prove that 4a2 – b2 + c2 + 4ac = 0.
Solution
It is given that
2a – b + c = 0
2a + c = b
By squaring on both sides
(2a + c)2 = b2
Expanding using formula
(2a)2 + (2×2a×c) + c2 = b2
By further calculation
4a2 + 4ac + c2 = b2
So we get
4a2 – b2 + c2 + 4ac = 0
Hence, it is proved.
27. If a + b + 2c = 0, prove that a3 + b3 + 8c3 = 6abc.
Solution
It is given that
a+b+2c = 0
We can write it as
a+b = –2c
By cubing on both sides
(a+b)3 = (-2c)3
Expanding using formula
a3+b3+ 3ab(a+b) = -8c3
Substituting the value of a + b
a3+b3+3ab(-2c) = -8c3
So we get
a3+b3+8c3 = 6abc
Hence, it is proved.
28. If a+b+c = 0, then find the value of a2/bc + b2/ca + c2/ab.
Solution
It is given that
a+b+c = 0
We can write it as
a3 + b3 + c3 – 3abc = 0
⇒ a3+b3+c3 = 3abc
Now dividing by abc on both sides
a3/abc + b3/abc + c3/abc = 3
By further calculation
a2/bc + b2/ac + c2/ab = 3
Therefore, the value of a2/bc + b2/ca + c2/ab is 3.
29. If x+y = 4, then find the value of x3+y3+12xy – 64.
Solution
It is given that
x + y = 4
By cubing on both sides
(x + y)3 = 43
Expanding using formula
x3 + y3 + 3xy(x + y) = 64
Substituting the value of x + y
x3 + y3 + 3xy(4) = 64
So we get
x3 + y3 + 12xy – 64 = 0
Hence, the value of x3 + y3 + 12xy – 64 is 0.
30. Without actually calculating the cubes, find the values of:
(i) (27)3 + (-17)3 + (-10)3
(ii) (-28)3 + (15)3 + (13)3
Solution
(i) (27)3 + (-17)3 + (-10)3
Consider a = 27, b = –17 and c = –10
We know that
a+b+c = 27–17–10 = 0
So a + b + c = 0
a3 + b3 + c3 = 3abc
Substituting the values
273 + (-17)3 + (-10)3 = 3(27)(-17)(-10)
= 13770
(ii) (-28)3 + (15)3 + (13)3
Consider a = –28, b = 15 and c = 13
We know that
a + b + c = –28+15+13 = 0
So a + b + c = 0
a3 + b3 + c3 = 3abc
Substituting the values
(-28)3 + (15)3 + (13)3 = 3(-28)(15)(13)
= –16380
31. Using suitable identity, find the value of:
Solution
Consider x = 86 and y = 14
= x + y
Substituting the values
= 86 + 14
= 100
Exercise 3.2
1. If x–y = 8 and xy = 5, find x2 + y2.
Solution
We know that
(x – y)2 = x2 + y2 – 2xy
It can be written as
x2 + y2 = (x – y)2 + 2xy
It is given that
x – y = 8 and xy = 5
Substituting the values
x2 + y2 = 82 + (2×5)
So we get
= 64 + 10
= 74
2. If x+y = 10 and xy = 21, find 2(x2+y2).
Solution
We know that
(x + y)2 = x2 + y2 + 2xy
It can be written as
x2 + y2 = (x + y)2 – 2xy
It is given that
(x + y) = 10 and xy = 21
Substituting the values
x2 + y2 = 102 – (2×21)
By further calculation
= 100 – 42
= 58
Here,
2 (x2 + y2) = 2×58 = 116
3. If 2a + 3b = 7 and ab = 2, find 4a2 + 9b2.
Solution
We know that
(2a + 3b)2 = 4a2 + 9b2 + 12ab
It can be written as
4a2 + 9b2 = (2a + 3b)2 – 12ab
It is given that
2a + 3b = 7
ab = 2
Substituting the values
4a2 + 9b2 = 72 –(12×2)
By further calculation
= 49 – 24
= 25
4. If 3x–4y = 16 and xy = 4, find the value of 9x2 + 16y2.
Solution
We know that
(3x – 4y)2 = 9x2 + 16y2 – 24xy
It can be written as
9x2 + 16y2 = (3x – 4y)2 + 24xy
It is given that
3x – 4y = 16 and xy = 4
Substituting the values
9x2 + 16y2= 162 + (24×4)
By further calculation
= 256 + 96
= 352
5. If x+y = 8 and x–y = 2, find the value of 2x2+2y2.
Solution
We know that
2 (x2 + y2) = (x + y)2+(x – y)2
It is given that
x + y = 8 and x – y = 2
Substituting the values
2x2 + 2y2 = 82 + 22
By further calculation
= 64 + 4
= 68
6. If a2 + b2 = 13 and ab = 6, find
(i) a + b
(ii) a – b
Solution
(i) We know that
(a + b)2 = a2 + b2 + 2ab
Substituting the values
= 13 + 2×6
So we get
= 13 + 12
= 25
Here
a + b = ±√25 = ±5
(ii) We know that
(a – b)2 = a2 + b2 – 2ab
Substituting the values
= 13 – 2×6
So we get
= 13 – 12
= 1
Here
a – b = ±√1
7. If a+b = 4 and ab = -12, find
(i) a – b
(ii) a2 – b2.
Solution
(i) We know that
(a – b)2 = a2 + b2 – 2ab
It can be written as
(a – b)2 = a2 + b2 + 2ab – 4ab
⇒ (a – b)2 = (a + b)2 – 4ab
It is given that
a + b = 4 and ab = –12
Substituting the values
(a – b)2 = 42 – 4(-12)
By further calculation
(a – b)2 = 16 + 48 = 64
So we get
(a – b) = ±√64 = ±8
(ii) We know that
a2 – b2 = (a+b) (a–b)
Substituting the values
a2 – b2 = (4×±8)
⇒ a2 – b2 = ± 32
8. If p–q = 9 and pq = 36, evaluate
(i) p + q
(ii) p2 – q2.
Solution
(i) We know that
(p+q)2 = p2+q2+2pq
It can be written as
(p+q)2 = p2+q2–2pq+4pq
⇒ (p+q)2 = (p–q)2+4pq
It is given that
p – q = 9 and pq = 36
Substituting the values
(p + q)2 = 92 + (4×36)
By further calculation
(p+q)2 = 81+144 = 225
So we get
p+q = ±√225 = ±15
(ii) We know that
p2–q2 = (p–q)(p+q)
Substituting the values
p2 – q2 = 9×±15
⇒ p2 – q2 = ±135
9. If x+y = 6 and x-y = 4, find
(i) x2 + y2
(ii) xy
Solution
(ii) We know that
(x+y)2 – (x–y)2 = 4xy
Substituting the values
62 – 42 = 4xy
By further calculation
36 – 16 = 4xy
⇒ 20 = 4xy
⇒ 4xy = 20
So we get
xy = 20/4 = 5
(i) x2 + y2 = (x+y)2 – 2xy
Substituting the values
= 62 – 2×5
By further calculation
= 36 – 10
= 26
10. If x–3 = 1/x, find the value of x2 + 1/x2.
Solution
It is given that
x – 3 = 1/x
We can write it as
x – 1/x = 3
Here
(x – 1/x)2 = x2 + 1/x2 – 2
So we get
x2 + 1/x2 = (x – 1/x)2 + 2
Substituting the values
x2 + 1/x2 = 32 + 2
By further calculation
= 9 + 2
= 11
11. If x + y = 8 and xy = 3¾, find the values of
(i) x – y
(ii) 3 (x2 + y2)
(iii) 5 (x2 + y2) + 4(x–y).
Solution
(i) We know that
(x – y)2 = x2 + y2 – 2xy
It can be written as
(x – y)2 = x2 + y2 + 2xy – 4xy
⇒ (x – y)2 = (x + y)2 – 4xy
It is given that
x + y = 8 and xy = 3¾ = 15/4
Substituting the values
(x – y)2 = 82 – (4× 15/4)
So we get
(x – y)2 = 65 – 15 = 49
⇒ x – y = ±√49 = ±7
(ii) We know that
(x + y)2 = x2 + y2 + 2xy
We can write it as
x2 + y2 = (x + y)2 – 2xy
It is given that
x + y = 8 and xy = 3¾ = 15/4
Substituting the values
x2 + y2 = 82 – (2× 15/4)
So we get
x2 + y2 = 64 – 15/2
Taking LCM
x2 + y2 = (128 – 15)/2 = 113/2
We get
3 (x2 + y2) = (3× 113/2) = 339/2 = 169½
(iii) We know that
5 (x2 + y2) + 4 (x–y) = (5 × 113/2) + (4×±7)
By further calculation
= 565/2 ± 28
We can write it as
= 565/2 + 28 or 565/2 – 28
= 621/2 or 509/2
It can be written as
= 310½ or 254½
12. If x2 + y2 = 34 and xy = 10 ½, find the value of 2 (x + y)2 + (x – y)2.
Solution
It is given that
x2 + y2 = 34 and xy = 10 ½ = 21/2
We know that
(x + y)2 = x2 + y2 + 2xy
Substituting the values
(x + y)2 = 34 + 2 (21/2)
So we get
(x + y)2 = 55 ….(1)
We know that
(x – y)2 = x2 + y2 – 2xy
Substituting the values
(x – y)2 = 34 – 2 (21/2)
So we get
(x – y)2 = 34 – 21 = 13 ….(2)
Using both the equations
2 (x + y)2 + (x – y)2 = (2×55) + 13 = 123
13. If a – b = 3 and ab = 4, find a3 – b3.
Solution
We know that
a3 – b3 = (a – b)3 + 3ab (a + b)
Substituting the values
a3 – b3 = 33 + (3×4×3)
By further calculation
a3 – b3 = 27 + 36 = 63
14. If 2a – 3b = 3 and ab = 2, find the value of 8a3 – 27b3.
Solution
We know that
8a3 – 27b3 = (2a)3 – (3b)3
According to the formula
= (2a – 3b)3 + (3×2a×3b)(2a – 3b)
By further simplification
= (2a – 3b)3 + 18ab (2a – 3b)
Substituting the values
= 33 + (18×2×3)
By further calculation
= 27 + 108
= 135
15. If x + 1/x = 4, find the values of
(i) x2 + 1/x2
(ii) x4 + 1/x4
(iii) x3 + 1/x3
(iv) x – 1/x.
Solution
(i) We know that
(x + 1/x)2 = x2 + 1/x2 + 2
It can be written as
x2 + 1/x2 = (x + 1/x)2 – 2
Substituting the values
= 42 – 2
= 16 – 2
= 14
(ii) We know that
(x2 + 1/x2)2 = x4 + 1/x4 + 2
It can be written as
x4 + 1/x4 = (x2 + 1/x2)2 – 2
Substituting the values
= 142 – 2
= 196 – 2
= 194
(iii) We know that
x3 + 1/x3 = (x + 1/x)3 – 3x(1/x)(x + 1/x)
It can be written as
(x + 1/x)3 – 3(x + 1/x) = 43 – (3×4)
By further calculation
= 64 – 12
= 52
(iv) We know that
(x – 1/x)2 = x2 + 1/x2 – 2
Substituting the values
= 14 – 2
= 12
So we get
x – 1/x = ±2√3
16. If x – 1/x = 5, find the value of x4 + 1/x4.
Solution
We know that
(x – 1/x)2 = x2 + 1/x2 – 2
It can be written as
x2 + 1/x2 = (x – 1/x)2 + 2
Substituting the values
x2 + 1/x2 = 52 + 2 = 27
Here
x4 + 1/x4 = (x2 + 1/x2)2 – 2
Substituting the values
x4 + 1/x4 = 272 – 2
So we get
= 729 – 2
= 727
17. If x – 1/x = √5, find the values of
(i) x2 + 1/x2
(ii) x + 1/x
(iii) x3 + 1/x3
Solution
(i) x2 + 1/x2 = (x – 1/x)2 + 2
Substituting the values
= (√5)2 + 2
= 5 + 2
= 7
(ii) (x + 1/x)2 = x2 + 1/x2 + 2
Substituting the values
= 7 + 2
= 9
Here
(x + 1/x)2 = 9
So we get
(x + 1/x) = ± √9 = ± 3
(iii) x3 + 1/x3 = (x + 1/x)3 – 3x(1/x)(x + 1/x)
Substituting the values
= (±3)3–3(±3)
By further calculation
= (±27) – (±9)
= ± 18
18. If x + 1/x = 6, find
(i) x – 1/x
(ii) x2 – 1/x2.
Solution
(i) We know that
(x – 1/x)2 = x2 + 1/x2 – 2
It can be written as
(x – 1/x)2 = x2 + 1/x2 + 2 – 4
⇒ (x – 1/x)2 = (x + 1/x)2 – 4
Substituting the values
(x – 1/x)2 = 62 – 4 = 32
So we get
x – 1/x = ±√32 = ±4√2
(ii) We know that
x2 – 1/x2 = (x – 1/x) (x + 1/x)
Substituting the values
x2 – 1/x2 = (±4√2)(6) = ±24√2
19. If x + 1/x = 2, prove that x2 + 1/x2 = x3 + 1/x3 = x4 + 1/x4.
Solution
We know that
x2 + 1/x2 = (x + 1/x) – 2
Substituting the values
x2 + 1/x2 = 22 – 2
So we get
x2 + 1/x2 = 4 – 2 = 2 ….(1)
⇒ x3 + 1/x3 = (x + 1/x)3 – 3 (x + 1/x)
Substituting the values
x3 + 1/x3 = 23 – 3 × 2
So we get
x3 + 1/x3 = 8 – 6 = 2 …(2)
⇒ x4 + 1/x4 = (x2 + 1/x2)2 – 2
Substituting the values
x4 + 1/x4 = 22 – 2
So we get
x4 + 1/x4 = 4 – 2 = 2 ….(3)
From equation (1), (2) and (3)
x2 + 1/x2 = x3 + 1/x3 = x4 + 1/x4
Hence, it is proved.
20. If x – 2/x = 3, find the value of x3 – 8/x3.
Solution
We know that
(x – 2/x)3 = x3 – 8/x3 – 3(x)(2/x)(x – 2/x)
By further simplification
(x – 2/x)3 = x3 – 8/x3 – 6(x – 2/x)
It can be written as
x3 – 8/x3 = (x – 2/x)3 + 6(x– 2/x)
Substituting the values
x3 – 8/x3 = 33 + (6×3)
By further calculation
x3 – 8/x3 = 27 + 18 = 45
21. If a + 2b = 5, prove that a3 + 8b3 + 30ab = 125.
Solution
We know that
(a + 2b)3 = a3 + 8b3 + 3(a)(2b)(a + 2b)
Substituting the values
53 = a3 + 8b3 + 6ab(5)
By further calculation
125 = a3 + 8b3 + 30ab
Therefore, a3 + 8b3 + 30ab = 125.
22. If a + 1/a = p, prove that a3 + 1/a3 = p(p2 – 3).
Solution
We know that
a3 + 1/a3 = (a + 1/a)3 – 3a(1/a)(a + 1/a)
Substituting the values
a3 + 1/a3 = p3 – 3p
Taking p as common
a3 + 1/a3 = p(p2 – 3)
Therefore, it is proved.
23. If x2+ 1/x2 = 27, find the value of x– 1/x.
Solution
We know that
(x – 1/x)2 = x2 + 1/x2 – 2
Substituting the values
(x – 1/x)2 = 27 – 2 = 25
So we get
x – 1/x = ±√25 = ±5
24. If x2 + 1/x2 = 27, find the value of 3x3+5x– 3/x3 – 5/x.
Solution
We know that
(x – 1/x)2 = x2 + 1/x2 – 2
Substituting the values
(x – 1/x)2 = 27 – 2 = 25
So we get
x – 1/x = ±√25 = ±5
Here
3x3 + 5x – 3/x3 – 5/x = 3(x3 – 1/x3) + 5(x – 1/x)
It can be written as
= 3 [(x – 1/x)3 + 3 (x – 1/x)] + 5 (x – 1/x)
Substituting the values
= 3 [(±5)3 + 3 (±5)] + 5 (±5)
By further calculation
= 3 [(±125) + (±15)] + (±25)
So we get
= (±375) + (±45) + (±25)
= ±445
25. If x2 + 1/25x2 = 8 3/5, find x + 1/5x.
Solution
We know that
(x + 1/5x)2 = x2 + 1/25x2 + 2x (1/5x)
It can be written as
(x + 1/5x)2 = x2 + 1/25x2 + 2/5
Substituting the values
(x + 1/5x)2 = 8 3/5 + 2/5
(x + 1/5x)2 = 43/5 + 2/5
So we get
(x + 1/5x)2 = 45/5 = 9
Here
x + 1/5x = ±√9 = ±3
26. If x2 + 1/4x2 = 8, find x3+ 1/8x3.
Solution
We know that
(x + 1/2x)2 = x2 + (1/2x)2 + 2x(1/2x)
It can be written as
(x + 1/2x)2 = x2 + 1/4x2 + 1
Substituting the values
(x + 1/2x)2 = 8 + 1 = 9
So we get
x + 1/2x = ±√9 = ±3
Here,
x3 + 1/8x3 = x3 + (1/2x)3
We know that
x3 + 1/8x3 = (x + 1/2x)3 – 3x(1/2x)(x + 1/2x)
Substituting the values
x3 + 1/8x3 = (±3)3 – 3/2(±3)
By further calculation
x3 + 1/8x3 = ±(27 – 9/2)
Taking LCM
x3 + 1/8x3 = ±(54–9)/2
⇒ x3 + 1/8x3 = ±45/2 = ±22 ½
Therefore, x3 + 1/8x3 = ±22 ½.
27. If a2–3a+1 = 0, find
(i) a2+ 1/a2
(ii) a3+ 1/a3.
Solution
It is given that
a2 – 3a + 1 = 0
By dividing each term by a
a + 1/a = 3
(i) We know that
(a + 1/a)2 = a2 + 1/a2 + 2
It can be written as
a2 + 1/a2 = (a + 1/a)2 – 2
Substituting the values
= 32 – 2
= 9 – 2
= 7
(ii) We know that
(a + 1/a)3 = a3 + 1/a3 + 3(a+ 1/a)
It can be written as
a3 + 1/a3 = (a + 1/a)3 – 3(a+ 1/a)
Substituting the values
= 33 – 3(3)
= 27 – 9
= 18
28. If a = 1/(a–5), find
(i) a – 1/a
(ii) a + 1/a
(iii) a2 – 1/a2.
Solution
It is given that
a = 1/ (a – 5)
We can write it as
a2 – 5a – 1 = 0
Now divide each term by a
a – 5 – 1/a = 0
So we get
a – 1/a = 5
(i) a – 1/a = 5
(ii) We know that
(a + 1/a)2 = (a – 1/a)2 + 4
Substituting the values
(a + 1/a)2 = 52 + 4
So we get
(a + 1/a)2 = 25 + 4 = 29
⇒ a + 1/a = ±√29
(iii) We know that
a2 – 1/a2 = (a + 1/a) (a – 1/a)
Substituting the values
a2 – 1/a2 = ±√29× 5
⇒ a2 – 1/a2 = ±5√29
29. If (x + 1/x)2 = 3, find x3 + 1/x3.
Solution
It is given that
(x + 1/x)2 = 3
⇒ (x + 1/x) = ±√3
We know that
x3 + 1/x3 = (x + 1/x)3 – 3(x + 1/x)
Substituting the values
x3 + 1/x3 = (±√3)3 – 3(±√3)
By further calculation
x3 + 1/x3 = (±3√3) – (±3√3) = 0
30. If x = 5 – 2√6, find the value of √x + 1/√x.
Solution
It is given that
x = 5 – 2√6
We can write it as
= 5 + 2√6
Here
x + 1/x = 5 – 2√6 + 5 + 2√6 = 10
So we get
(√x + 1/√x)2 = x + 1/x + 2
Substituting the values
= 10 + 2
= 12
31. If a+b+c = 12 and ab+bc+ca = 22, find a2+b2+c2.
Solution
We know that
(a+b+c)2 = a2+b2+c2 + 2(ab+bc+ca)
We can write it as
a2+b2+c2 = (a+b+c)2 – 2 (ab+bc+ca)
Substituting the values
a2+b2+c2 = 122 – 2(22)
By further calculation
a2+b2+c2 = 144 – 44 = 100
32. If a+b+c = 12 and a2+b2+c2 = 100, find ab+bc+ca.
Solution
We know that
(a+b+c)2 = a2+b2+c2+2ab+2bc+2ca
It can be written as
2ab+2bc+2ca = (a+b+c)2 – (a2+b2+c2)
Taking out 2 as common
2(ab+bc+ca) = 122 –100 = 144–100 = 44
By further calculation
ab+bc+ca = 44/2 = 22
33. If a2+b2+c2 = 125 and ab+bc+ca = 50, find a+b+c.
Solution
We know that
(a+b+c)2 = a2+b2+c2 + 2(ab+bc+ca)
Substituting the values
(a+b+c)2 = 125 + 2(50)
By further calculation
(a+b+c)2 = 125 + 100 = 225
So we get
a+b+c = ±√225 = ±15
34. If a+b–c = 5 and a2+b2+c2 = 29, find the value of ab–bc–ca.
Solution
It is given that
a+b–c = 5
By squaring on both sides
(a+b–c)2 = 52
Expanding using formula
a2+b2+c2+2ab–2bc–2ca = 25
Substituting the values and taking 2 as common
29 + 2(ab–bc–ca) = 25
By further calculation
2(ab–bc–ca) = 25 – 29 = – 4
So we get
ab – bc – ca = – 4/2 = –2
Therefore, ab – bc – ca = –2.
35. If a – b = 7 and a2 + b2 = 85, then find the value of a3 – b3.
Solution
We know that
(a–b)2 = a2 + b2 – 2ab
Substituting the values
72 = 85 – 2ab
By further calculation
49 = 85–2ab
So we get
2ab = 85 – 49 = 36
Dividing by 2
ab = 36/2 = 18
Here,
a3 – b3 = (a–b)(a2+b2+ab)
Substituting the values
a3 – b3 = 7(85+18)
By further calculation
a3 – b3 = 7×103
So we get
a3 – b3 = 721
36. If the number x is 3 less than the number y and the sum of the squares of x and y is 29, find the product of x and y.
Solution
It is given that
x = y – 3 and x2 + y2 = 29
It can be written as
x – y = –3
By squaring on both sides
(x – y)2 = (-3)2
Expanding using formula
x2 + y2 – 2xy = 9
Substituting the values
29 – 2xy = 9
By further calculation
-2xy = 9–29 = –20
Dividing by 2
xy = –20/-2 = 10
So we get
xy = 10
37. If the sum and the product of two numbers are 8 and 15 respectively, find the sum of their cubes.
Solution
Consider x and y as the two numbers
x + y = 8 and xy = 15
By cubing on both sides
(x + y)3 = 83
Expanding using formula
x3 + y3 + 3xy(x+y) = 512
Substituting the values
x3 + y3 + (3×15×8) = 512
By further calculation
x3 + y3 + 360 = 512
So we get
x3 + y3 = 512 – 360 = 152
Chapter Test
1. Find the expansions of the following :
(i) (2x + 3y + 5) (2x + 3y – 5)
(ii) (6 – 4a -7b)2
(iii) (7 – 3xy)3
(iv) (x + y + 2)3
Solution
(i) (2x + 3y + 5) (2x + 3y – 5)
Let us simplify the expression, we get
(2x + 3y + 5) (2x + 3y – 5) = [(2x + 3y) + 5] [(2x – 3y) – 5]
By using the formula, (a)2 – (b)2 = [(a + b) (a – b)]
= (2x + 3y)2 – (5)2
= (2x)2 + (3y) 2 + (2×2x×3y) – (5×5)
= 4x2 + 9y2 + 12xy – 25
(ii) (6–4a–7b)2
Let us simplify the expression, we get
(6–4a–7b)2 = [6+(-4a)+(-7b)]2
= (6)2 + (-4a)2 + (-7b)2 + 2(6)(-4a) + 2(-4a)(-7b) + 2(-7b)(6)
= 36 + 16a2 + 49b2 – 48a + 56ab – 84b
(iii) (7 – 3xy)3
Let us simplify the expression
By using the formula, we get
(7 – 3xy)3 = (7)3 – (3xy)3 – 3(7)(3xy)(7–3xy)
= 343 – 27x3y3 – 63xy(7–3xy)
= 343 – 27x3y3 – 441xy + 189x2y2
(iv) (x+y+2)3
Let us simplify the expression
By using the formula, we get
(x+y+2)3 = [(x+y)+2]3
= (x + y)3 + (2)3 + 3(x+y)(2)(x+y+2)
= x3 + y3 + 3x2y + 3xy2 + 8 + 6(x+y)[(x+y)+2]
= x3 + y3 + 3x2y + 3xy2 + 8 + 6(x+y)2 +12(x+y)
= x3 + y3 + 3x2y + 3xy2 + 8 + 6(x2+y2+2xy) + 12x + 12y
= x3 + y3 + 3x2y + 3xy2 + 8 + 6x2 + 6y2 + 12xy + 12x + 12y
= x3 + y3 + 3x2y + 3xy2 + 8 + 6x2 + 6y2 + 12x + 12y + 12xy
2. Simplify: (x–2)(x+2)(x2+4)(x4+16)
Solution
Let us simplify the expression, we get
(x–2)(x+2)(x4+4)(x4+16)
= (x2–4)(x4+4)(x4+16)
= [(x2)2–(4)2](x4+16)
= (x4–16)(x4+16)
= (x4)2 – (16)2
= x8 – 256
3. Evaluate 1002×998 by using a special product.
Solution
Let us simplify the expression, we get
1002×998
= (1000+2)(1000–2)
= (1000)2 – (2)2
= 1000000 – 4
= 999996
4. If a+2b+3c = 0, Prove that a3+8b3+27c3 = 18abc
Solution
Given:
a+2b+3c = 0, a+2b = –3c
Let us cube on both the sides, we get
(a+2b)3 = (-3c)3
⇒ a3 + (2b)3 + 3(a) (2b) (a + 2b) = -27c3
⇒ a3 + 8b3 + 6ab (– 3c) = – 27c3
⇒ a3 + 8b3 – 18abc = -27c3
⇒ a3 + 8b3 + 27c3 = 18abc
Hence, proved.
5. If 2x = 3y – 5, then find the value of 8x3 – 27y3 + 90xy + 125.
Solution
Given:
2x = 3y – 5
Now, let us cube on both sides, we get
(2x – 3y)3 = (-5)3
⇒ (2x)3 – (3y)3 – (3×2x×3y)(2x – 3y) = -125
⇒ 8x3 – 27y3 – 18xy(2x – 3y) = -125
Now, substitute the value of 2x – 3y = -5
8x3 – 27y3 – 18xy (-5) = -125
⇒ 8x3 – 27y3 + 90xy = -125
⇒ 8x3 – 27y3 + 90xy + 125 = 0
6. If a2 – 1/a2 = 5, evaluate a4 + 1/a4
Solution
It is given that,
a2 – 1/a2 = 5
So,
By using the formula, (a + b)2
[a2 – 1/a2]2 = a4 + 1/a4 – 2
⇒ [a2 – 1/a2]2 + 2 = a4 + 1/a4
Substitute the value of a2 – 1/a2 = 5, we get
52 + 2 = a4 + 1/a4
⇒ a4 + 1/a4 = 25 + 2 = 27
7. If a + 1/a = p and a – 1/a = q, Find the relation between p and q.
Solution
It is given that,
a + 1/a = p and a – 1/a = q
So,
(a + 1/a)2 – (a – 1/a)2 = 4(a)(1/a)
= 4
By substituting the values, we get
p2 – q2 = 4
Hence the relation between p and q is that p2 – q2 = 4.
8. If (a2 + 1)/a = 4, find the value of 2a3 + 2/a3
Solution
It is given that,
(a2 + 1)/a = 4
⇒ a2/a + 1/a = 4
⇒ a + 1/a = 4
So by multiplying the expression by 2a, we get
2a3 + 2/a3 = 2[a3 + 1/a3]
= 2[(a + 1/a)3 – 3(a)(1/a)(a + 1/a)]
= 2[(4)3 – 3(4)]
= 2[64 – 12]
= 2(52)
= 104
9. If x = 1/(4–x), find the value of
(i) x + 1/x
(ii) x3 + 1/x3
(iii) x6 + 1/x6
Solution
It is given that,
x = 1/(4–x)
So,
(i) x(4–x) = 1
4x – x2 = 1
Now let us divide both sides by x, we get
4 – x = 1/x
⇒ 4 = 1/x + x
⇒1/x + x = 4
⇒ 1/ x + x = 4
(ii) x3 + 1/x3 = (x + 1/x)2 – 3(x + 1/x)
By substituting the values, we get
= (4)3 –3(4)
= 64 – 12
= 52
(iii) x6 + 1/x6 = (x3 + 1/x3)2 – 2
= (52)2 – 2
= 2704 – 2
= 2702
10. If x – 1/x = 3 + 2√2, find the value of ¼(x3 – 1/x3)
Solution
It is given that,
x – 1/x = 3 + 2√2
So,
x3 – 1/x3 = (x – 1/x)3 + 3(x – 1/x)
= (3 + 2√2)3 + 3(3+2√2)
By using the formula, (a+b)3 = a3 + b3 + 3ab(a+b)
= (3)3 + (2√2)3 + 3(3)(2√2)(3+2√2) + 3(3+2√2)
= 27 + 16√2 + 54√2 + 72 + 9 + 6√2
= 108 + 76√2
Hence,
¼ (x3 – 1/x3) = ¼ (108+76√2)
= 27 + 19√2
11. If x + 1/x = 31⁄3, find the value of x3 – 1/x3
Solution
It is given that,
x + 1/x = 31⁄3
Also, 31⁄3 = 10/3
we know that,
(x – 1/x)2 = x2 + 1/x2 – 2
= x2 + 1/x2 + 2 – 4
= (x + 1/x)2 – 4
But x + 1/x = 10/3
So,
(x – 1/x)2 = (10/3)2 – 4
= 100/9 – 4
= (100 – 36)/9
= 64/9
x – 1/x = √(64/9)
= 8/3
Now,
x3 – 1/x3 = (x– 1/x)3 + 3(x)(1/x)(x– 1/x)
= (8/3)3 + 3(8/3)
= (512/27) + 8
= 728/27
12. If x = 2–√3, then find the value of x3 – 1/x3
Solution
It is given that,
x = 2 – √3
so,
1/x = 1/(2–√3)
By rationalizing the denominator, we get
= [1(2+√3)]/[(2–√3)(2+√3)]
= [(2+√3)]/[(22) – (√3)2]
= [(2+√3)]/(4–3)
= 2+√3
Now,
x – 1/x = 2–√3–2–√3
= –2√3
Let us cube on both sides, we get
(x – 1/x)3 = (-2√3)3
⇒ x3 – 1/x3 – 3(x)(1/x)(x – 1/x) = 24√3
⇒ x3 – 1/x3 – 3(-2√3) = -24√3
⇒ x3 – 1/x3 + 6√3 = -24√3
⇒ x3 – 1/x3 = -24√3 – 6√3
= -30√3
Hence,
x3 – 1/x3 = -30√3
13. If the sum of two numbers is 11 and sum of their cubes is 737, find the sum of their squares.
Solution
Let us consider x and y be two numbers
Then,
x + y = 11
x3 + y3 = 735 and x2 + y2 =?
Now,
x + y = 11
Let us cube on both the sides,
(x + y)3 = (11)3
⇒ x3 + y3 + 3xy(x + y) = 1331
⇒ 737 + (3x×11) = 1331
⇒ 33xy = 1331 – 737
= 594
⇒ xy = 594/33
⇒ xy = 8
We know that, x + y = 11
By squaring on both sides, we get
(x + y)2 = (11)2
⇒ x2 + y2 + 2xy = 1212x2 + y2 + (2×18) = 121
⇒ x2 + y2 + 36 = 121
⇒ x2 + y2 = 121 – 36
= 85
Hence sum of the squares = 85
14. If a – b = 7 and a3 – b3 = 133, find:
(i) ab
(ii) a2 + b2
Solution
It is given that,
a – b = 7
let us cube on both sides, we get
(i) (a – b)3 = (7)3
a3 + b3 – 3ab(a–b) = 343
⇒ 133 – (3ab×7) = 343
⇒ 133 – 21ab = 343
⇒ –21ab = 343 – 133 = 210
⇒ ab = -210/21
⇒ ab = -10
(ii) a2 + b2
Again a – b = 7
Let us square on both sides, we get
(a – b)2 = (7)2
a2 + b2 – 2ab = 49
a2 + b2 – 2×(-10) = 49
a2 + b2 + 20 = 49
a2 + b2 = 49 – 20
= 29
Hence, a2 + b2 = 29
15. Find the coefficient of x2 expansion of (x2 + x + 1)2 + (x2 – x + 1)2
Solution
Given:
The expression, (x2 + x + 1)2 + (x2 – x + 1)2
(x2 + x + 1)2 + (x2 – x + 1)2 = [((x2 + 1) + x)2 + [(x2 + 1)– x)2]
= (x2 + 1)2 + x2 + 2 (x2 + 1) (x) + (x2 + 1)2 + x2 – 2 (x2 + 1) (x)
= (x2)2 + (1)2 + (2×x2×1) + x2 + (x2)2 + 1 + (2×x2) + 1 + x2
= x4 + 1 + 2x2 + x2 + x4 + 1 + 2x2 + x2
= 2x4 + 6x2 + 2
∴ Co-efficient of x2 is 6.