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Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Exercise 4(A) Page: 57 1. Find the square of:
(i) 2a+b
(ii) 3a+7b
(iii) 3a-4b
(iv) 𝟑𝐚
𝟐𝐛−
𝟐𝐛
𝟑𝐚
Solution:
(i)
(ii)
(iii)
(iv)
2. Use identities to evaluate:
(i) 1012
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(ii) 5022
(iii) 972
(iv) 9982
Solution: (i)
(ii)
(iii)
(iv)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
3. Evaluate: (i)
(ii)
Solution:
(i)
(ii)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
4. Evaluate: (i)
(ii)
Solution:
(i)Consider the given expression:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(ii)Consider the given expression:
5. If a+b=7 and ab = 10; find a-b.
Solution:
6. If a-b=7 and ab = 18; find a+b.
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Solution:
7. If 𝒙 + 𝒚 =𝟕
𝟐 and 𝒙𝒚 =
𝟓
𝟐; find:
(i) x-y
(ii) x2-y2
Solution: (i)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(ii)
8. If a-b=0.9 and ab = 0.36; find
(i) a+b.
(ii) a2-b2
Solution: (i)
(ii)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
9. If a-b=4 and a+b=6; find :
(i) a2+b2
(ii) ab
Solution: (i)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(ii)
10. If 𝒂 +𝟏
𝒂= 𝟔 and 𝒂 ≠ 0; find:
(i) 𝒂 −𝟏
𝒂
(ii) 𝒂𝟐 −𝟏
𝒂𝟐
Solution: (i)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(ii)
11. If 𝒂 −𝟏
𝒂= 𝟖 and 𝒂 ≠ 0; find:
(i) 𝒂 +𝟏
𝒂
(ii) 𝒂𝟐 −𝟏
𝒂𝟐
Solution:
(i)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(ii)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
12. If a2-3a+1=0 and a≠0; find:
(i) 𝒂 +𝟏
𝒂
(ii) 𝒂𝟐 +𝟏
𝒂𝟐
Solution: (i)
(ii)
13. If a2-5a-1=0 and a≠0; find:
(i) 𝒂 −𝟏
𝒂
(ii) 𝒂 +𝟏
𝒂
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(iii) 𝒂𝟐 −𝟏
𝒂𝟐
Solution: (i)
(ii)
(iii)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
14. If 3x+4y=16 and xy=4; find the value of 9x2+16y2
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
15. The number x is 2 more than the number y. If the sum of the squares of x and
y is 34; find the product of x and y.
Solution: Given x is 2 more than y, so x = y + 2
Sum of squares of x and y is 34, so x2 + y2 = 34.
Replace x = y + 2 in the above equation and solve for y.
We get (y + 2)2 + y2 = 34
2y2 + 4y - 30 = 0
y2 + 2y - 15 = 0
(y + 5)(y - 3) = 0
So y = -5 or 3
For y = -5, x =-3
For y = 3, x = 5
Product of x and y is 15 in both cases.
16. The difference between two positive numbers is 5 and the sum of their
squares is 73. Find the product of these numbers.
Solution: Let the two positive numbers be a and b.
Given difference between them is 5 and sum of squares is 73.
So, a - b = 5, a2 + b2 = 73
Squaring on both sides gives
(a - b)2 = 52
a2 + b2 - 2ab = 25
but a2 + b2 = 73
so 2ab = 73 - 25 = 48
ab = 24
So, the product of numbers is 24.
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Exercise 4(B) Page: 60
1. Find the cube of:
(i) 3a-2b
(ii) 5a+3b
(iii) 𝟐𝒂 +𝟏
𝟐𝒂 ; (𝒂 ≠ 𝟎)
(iv) 𝟑𝒂 −𝟏
𝒂 ; (𝒂 ≠ 𝟎)
Solution: (i)
(ii)
(iii)
(iv)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
2. If 𝒂𝟐 +𝟏
𝒂𝟐= 𝟒𝟕 and 𝒂 ≠ 0; find:
(i) 𝒂 +𝟏
𝒂
(ii) 𝒂𝟑 +𝟏
𝒂𝟑
Solution: (i)
(ii)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
3. If 𝒂𝟐 +𝟏
𝒂𝟐= 𝟏𝟖 and 𝒂 ≠ 0; find:
(i) 𝒂 −𝟏
𝒂
(ii) 𝒂𝟑 −𝟏
𝒂𝟑
Solution:
(i)
(ii)
4. If 𝒂 +𝟏
𝒂= 𝒑 and 𝒂 ≠ 0; then show that:
𝒂𝟑 +𝟏
𝒂𝟑= 𝒑(𝒑𝟐 − 𝟑)
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
5. If a + 2b = 5; then show that:
a3+8b3+30ab=125
Solution:
6. If (𝒂 +𝟏
𝒂)
𝟐= 𝟑 𝒂𝒏𝒅 𝒂 ≠ 𝟎; then show that:
𝒂𝟑 +𝟏
𝒂𝟑= 𝟎
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
7. If a+2b+c=0; then show that:
a3+8b3+c3=6abc.
Solution:
8. Use property to evaluate:
(i) 133 + (-8)3 + (-5)3
(ii) 73 + 33 + (-10)3
(iii) 93 - 53 - 43
(iv) 383 + (-26)3 + (-12)3
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Solution: Property is if a + b + c = 0 then a3 + b3 + c3 = 3abc
(i) a = 13, b = -8 and c = -5
133 + (-8)3 + (-5)3 = 3(13)(-8)(-5) = 1560
(ii) a = 7, b = 3, c = -10
73 + 33 + (-10)3 = 3(7)(3)(-10) = -630
(iii)a = 9, b = -5, c = -4
93 - 53 - 43 = 93 + (-5)3 + (-4)3 = 3(9)(-5)(-4) = 540
(iv) a = 38, b = -26, c = -12
383 + (-26)3 + (-12)3 = 3(38)(-26)(-12) = 35568
9. If 𝒂 ≠ 0 and 𝒂 −𝟏
𝒂= 𝟑; find:
(i) 𝒂𝟐 +𝟏
𝒂𝟐
(ii) 𝒂𝟑 −𝟏
𝒂𝟑
Solution: (i)
(ii)
10. If 𝒂 ≠ 0 and 𝒂 −𝟏
𝒂= 𝟒; find:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(i) 𝒂𝟐 +𝟏
𝒂𝟐
(ii) 𝒂𝟒 +𝟏
𝒂𝟒
(iii) 𝒂𝟑 −𝟏
𝒂𝟑
Solution: (i)
(ii)
(iii)
11. If 𝒙 ≠ 0 and 𝒙 −𝟏
𝒙= 𝟐; then show that:
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Thus from equations (1), (2) and (3), we have
12. If 2x-3y=10 and xy=16; find the value of 8x3-27y3.
Solution: Given that 2x - 3y = 10, xy = 16
:. (2x - 3y)3 = (10)3
Þ 8x3 - 27y3 - 3 (2x) (3y) (2x - 3y) = 1000 Þ 8x3 - 27 y3 -18xy (2x - 3y) = 1000
Þ 8x3 - 27 y3 - 18 (16) (10) = 1000
Þ 8x3 - 27 y3 - 2880 = 1000
Þ 8x3 - 27 y3 = 1000 + 2880
Þ8x3 - 27 y3 =3880
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
13. Expand:
(i) (3x + 5y + 2z) (3x - 5y + 2z)
(ii) (3x - 5y - 2z) (3x - 5y + 2z)
Solution: (i)
(3x + 5y + 2z) (3x - 5y + 2z)
= {(3x + 2z) + (5y)} {(3x + 2z) - (5y)}
= (3x + 2z)2 - (5y)2
{since (a + b) (a - b) = a2 - b2}
= 9x2 + 4z2 + 2 × 3x × 2z - 25y2
= 9x2 + 4z2 + 12xz - 25y2
= 9x2 + 4z2 - 25y2 + 12xz
(ii)
(3x - 5y - 2z) (3x - 5y + 2z)
= {(3x - 5y) - (2z)} {(3x - 5y) + (2z)}
= (3x - 5y)2 - (2z)2{since(a + b) (a - b) = a2 - b2}
= 9x2 + 25y2 - 2 × 3x × 5y - 4z2
= 9x2 + 25y2- 30xy - 4z2
= 9x2 +25y2 - 4z2 - 30xy
14. The sum of two numbers is 9 and their product is 20. Find the sum of their:
(i) Squares
(ii) Cubes
Solution: Given sum of two numbers is 9 and their product is 20.
Let the numbers be a and b.
a + b = 9
ab = 20
(i) Squaring on both sides gives
(a+b)2 = 92
a2 + b2 + 2ab = 81
a2 + b2 + 40 = 81
So sum of squares is 81 - 40 = 41
(ii) Cubing on both sides gives
(a + b)3 = 93
a3 + b3 + 3ab(a + b) = 729
a3 + b3 + 60(9) = 729
a3 + b3 = 729 - 540 = 189
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
So the sum of cubes is 189.
15. Two positive numbers x and y are such that x>y. if the difference of these
numbers is 5 and their product is 24, find:
(i) Sum of these numbers.
(ii) Difference of their cubes.
(iii) Sum of their cubes.
Solution: (i) Given x - y = 5 and xy = 24 (x>y)
(x + y)2 = (x - y)2 + 4xy = 25 + 96 = 121
So, x + y = 11; sum of these numbers is 11.
(ii) Difference of their Cubes
(x - y)3 = 53
x3 - y3 - 3xy(x - y) = 125
x3 - y3 - 72(5) = 125
x3 - y3= 125 + 360 = 485
So, difference of their cubes is 485.
(iii) Difference of their Cubes
(x + y)3 = 113
x3 + y3 + 3xy(x + y) = 1331
x3 + y3 = 1331 - 72(11) = 1331 - 792 = 539
So, sum of their cubes is 539.
16. If 4x2+y2=a and xy=b, find the value of 2x+y
Solution: xy = ab ….(i)
4x2 + y2 = a ….(ii)
Now, (2x + y)2 = (2x)2 + 4xy + y2
= 4x2 + y2 + 4xy
= a + 4b ….[From (i) and (ii)]
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Exercise 4(C) Page: 62 1. Expand:
(i) (x+8)(x+10)
(ii) (x+8)(x-10)
(iii) (x-8)(x+10)
(iv) (x-8)(x-10)
Solution:
2. Expand:
(i) (𝟐𝒙 −𝟏
𝒙) (𝟑𝒙 +
𝟐
𝒙)
(ii) (𝟑𝒂 +𝟐
𝒃) (𝟐𝒂 −
𝟑
𝒃)
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
3. Expand:
Solution:
4. If a+b+c=12 and a2+b2+c2=50; find ab+bc+ca.
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
5. If a2+b2+c2=35 and ab+bc+ca=23; find a+b+c.
Solution:
6. If a+b+c=p and ab+bc+ca=q; find a2+b2+c2.
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
7. If a2+b2+c2=50 and ab+bc+ca=47; find a+b+c.
Solution:
8. If x+y-z=4 and x2+y2+z2=30, then find the value of xy-yz-zx.
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Exercise 4(D) Page: 64
1. If x+2y+3z=0 and
x3+4y3+9z3=18xyz; evaluate;
Solution:
Given that x3 + 4y3 + 9z3 = 18xyzand x + 2y + 3z = 0
x + 2y = - 3z, 2y + 3z = -x and 3z + x = -2y
Now
2. If 𝒂 +𝟏
𝒂= 𝒎 and 𝒂 ≠ 𝟎; find in terms of ‘m’; the value of
(i) 𝒂 −𝟏
𝒂
(ii) 𝒂𝟐 −𝟏
𝒂𝟐
Solution: (i)
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(ii)
3. In the expansion of (2x2-8)(x-4)2; find the value of:
(i) Coefficient of x3
(ii) Coefficient of x2
(iii) Constant term.
Solution:
4. If x>0 and 𝒙𝟐 +𝟏
𝟗𝒙𝟐=
𝟐𝟓
𝟑𝟔, find: 𝒙𝟑 +
𝟏
𝟐𝟕𝒙𝟑.
Solution: Given that
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
5. If 2(x2+1)=5x, find:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(i) 𝒙 −𝟏
𝒙
(ii) 𝒙𝟑 −𝟏
𝒙𝟑
Solution: (i)
2(x2 + 1} = 5x
Dividing by x, we have
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
(ii)
6. If a2 + b2 = 34 and ab= 12; find:
(i) 3(a + b)2 + 5(a - b)2
(ii) 7(a - b)2 - 2(a + b)2
Solution: a2 + b2 = 34, ab= 12
(a + b)2 = a2 + b2 + 2ab
= 34 + 2 x 12 = 34 + 24 = 58
(a - b)2 = a2 + b2 - 2ab
= 34 - 2 x 12 = 34- 24 = 10
(i) 3(a + b)2 + 5(a - b)2
= 3 x 58 + 5 x 10 = 174 + 50
= 224
(ii) 7(a - b)2 - 2(a + b)2
= 7 x 10 - 2 x 58 = 70 - 116 = -46
7. If 𝟑𝒙 −𝟒
𝒙 =4 and 𝒙 ≠ 𝟎; find:
Solution:
Given 3x -
We need to find
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
8. If and x ≠0; find the value of:
Solution:
Given that
We need to find the value of
Consider the given equation:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
9. If and x ≠5, find: 𝒙𝟐 −𝟏
𝒙𝟐
Solution:
By cross multiplication,
=> x (x - 5) = 1 => x2 - 5x = 1 => x2 - 1 = 5x
Dividing both sides by x,
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
10. If and x ≠5, find: 𝒙𝟑 −𝟏
𝒙𝟑
Solution:
By cross multiplication,
=> x (5 - x) = 1 => x2 - 5x =-1 => x2 + 1 = 5x
Dividing both sides by x,
11. If 3a + 5b + 4c = 0, show that:
27a3 + 125b3 + 64c3 = 180abc
Solution: Given that 3a + 5b + 4c = 0
3a + 5b = -4c
Cubing both sides,
(3a + 5b)3 = (-4c)3
=>(3a)3 + (5b)3 + 3 x 3a x 5b (3a + 5b) = -64c3
=>27a3 + 125b3 + 45ab x (-4c) = -64c3
=>27a3 + 125b3 - 180abc = -64c3
=>27a3 + 125b3 + 64c3 = 180abc
Hence proved.
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
12. The sum of two numbers is 7 and the sum of their cubes is 133, find the sum
of their squares.
Solution: Let a, b be the two numbers
.'. a + b = 7 and a3 + b3 = 133
(a + b)3 = a3 + b3 + 3ab (a + b)
=> (7)3 = 133 + 3ab (7)
=> 343 = 133 + 21ab => 21ab = 343 - 133 = 210
=> 21ab = 210 => ab= 2I
Now a2 + b2 = (a + b)2 - 2ab
=72 - 2 x 10 = 49 - 20 = 29
13. In each of the following, find the value of ‘a’:
(i) 4x2 + ax + 9 = (2x + 3)2
(ii) 4x2 + ax + 9 = (2x - 3)2
(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2
Solution: (i) 4x2 + ax + 9 = (2x + 3)2
Comparing coefficients of x terms, we get
ax = 12x
so, a = 12
(ii) 4x2 + ax + 9 = (2x - 3)2
Comparing coefficients of x terms, we get
ax = -12x
so, a = -12
(iii) 9x2 + (7a - 5)x + 25 = (3x + 5)2
Comparing coefficients of x terms, we get
(7a - 5)x = 30x
7a - 5 = 30
7a = 35
a = 5
14. If 𝒙𝟐+𝟏
𝒙= 𝟑
𝟏
𝟑 and x>1; find
(i) 𝒙 −𝟏
𝒙
(ii) 𝒙𝟑 −𝟏
𝒙𝟑
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Solution: Given
15. The difference between two positive numbers is 4 and the difference between
their cubes is 316. Find:
(i) their product
(ii) the sum of their squares.
Solution:
Given difference between two positive numbers is 4 and difference
between their cubes is 316.
Let the positive numbers be a and b
a - b = 4
a3 - b3 = 316
Cubing both sides,
(a - b)3 = 64
a3 - b3 - 3ab(a - b) = 64
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Given a3 - b3 = 316
So 316 - 64 = 3ab(4)
252 = 12ab
So ab = 21; product of numbers is 21
Squaring both sides, we get
(a - b)2 = 16
a2 + b2 - 2ab = 16
a2 + b2 = 16 + 42 = 58
Sum of their squares is 58.
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Exercise 4(E) Page: 66 1. Simplify:
(i) (x + 6)(x + 4)(x - 2)
(ii) (x - 6)(x - 4)(x + 2)
(iii) (x - 6)(x - 4)(x - 2)
(iv) (x + 6)(x - 4)(x - 2)
Solution: Using identity:
(x + a)(x + b)(x + c) = x3 + (a + b + c)x2 + (ab + bc + ca)x + abc
(i) (x + 6)(x + 4)(x - 2)
= x3 + (6 + 4 - 2)x2 + [6 × 4 + 4 × (-2) + (-2) × 6]x + 6 × 4 × (-2)
= x3 + 8x2 + (24 - 8 - 12)x - 48
= x3 + 8x2 + 4x - 48
(ii) (x - 6)(x - 4)(x + 2)
= x3 + (-6 - 4 + 2)x2 + [-6 × (-4) + (-4) × 2 + 2 × (-6)]x + (-6) × (-4) × 2
= x3 - 8x2 + (24 - 8 - 12)x + 48
= x3 - 8x2 + 4x + 48
(iii) (x - 6)(x - 4)(x - 2)
= x3 + (-6 - 4 - 2)x2 + [-6 × (-4) + (-4) × (-2) + (-2) × (-6)]x + (-6) × (-4) × (-2)
= x3 - 12x2 + (24 + 8 + 12)x - 48
= x3 - 12x2 + 44x - 48
(iv) (x + 6)(x - 4)(x - 2)
= x3 + (6 - 4 - 2)x2 + [6 × (-4) + (-4) × (-2) + (-2) × 6]x + 6 × (-4) × (-2)
= x3 - 0x2 + (-24 + 8 - 12)x + 48
= x3 - 28x + 48
2. Simplify using following identity: (𝑎 ± 𝑏)(𝑎2 ± 𝑎𝑏 + 𝑏2) = 𝑎3 ± 𝑏3
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
3. Using suitable identity, evaluate:
(i) (104)3
(ii) (97)3
Solution: Using identity: (a ± b)3 = a3 ± b3 ± 3ab(a ± b)
(i) (104)3 = (100 + 4)3
= (100)3 + (4)3 + 3 × 100 × 4(100 + 4)
= 1000000 + 64 + 1200 × 104
= 1000000 + 64 + 124800
= 1124864
(ii) (97)3 = (100 - 3)3
= (100)3 - (3)3 - 3 × 100 × 3(100 - 3)
= 1000000 - 27 - 900 × 97
= 1000000 - 27 - 87300
= 912673
4. Simplify:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
Solution:
5. Evaluate:
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
6. If a - 2b + 3c = 0, state the value of a3 - 8b3 + 27c3
Solution: a3 - 8b3 + 27c3 = a3 + (-2b)3 + (3c)3
Since a - 2b + 3c = 0, we have
a3 - 8b3 + 27c3 = a3 + (-2b)3 + (3c)3
= 3(a)( -2b)(3c)
= -18abc
7. If x + 5y = 10; find the value of x3 + 125y3 + 150xy - 1000.
Solution: x + 5y = 10
⇒ (x + 5y)3 = 103
⇒ x3 + (5y)3 + 3(x)(5y)(x + 5y) = 1000
⇒ x3 + (5y)3 + 3(x)(5y)(10) = 1000
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
= x3 + (5y)3 + 150xy = 1000
= x3 + (5y)3 + 150xy - 1000 = 0
8. If , find:
Solution:
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
9. If ; find a3+b3.
Solution:
10. Prove that:
x2 + y2 + z2 - xy - yz – zx is always positive.
Solution: x2 + y2 + z2 - xy - yz - zx
= 2(x2 + y2 + z2 - xy - yz - zx)
= 2x2 + 2y2 + 2z2 - 2xy - 2yz - 2zx
= x2 + x2 + y2 + y2 + z2 + z2 - 2xy - 2yz - 2zx
= (x2 + y2 - 2xy) + (z2 + x2 - 2zx) + (y2 + z2 - 2yz)
= (x - y)2 + (z - x)2 + (y - z)2
Since square of any number is positive, the given equation is always positive.
11. Find:
(i) (a + b)(a + b) = (a + b)2
(ii) (a + b)(a + b)(a + b)
(iii) (a - b)(a - b)(a - b) by using the result of part (ii)
Solution: (i) (a + b)(a + b) = (a + b)2
= a × a + a × b + b × a + b × b
Concise Selina Solutions for Class 9 Maths Chapter 4-
Expansions
= a2 + ab + ab + b2
= a2 + b2 + 2ab
(ii) (a + b)(a + b)(a + b)
= (a × a + a × b + b × a + b × b)(a + b)
= (a2 + ab + ab + b2)(a + b)
= (a2 + b2 + 2ab)(a + b)
= a2 × a + a2 × b + b2 × a + b2 × b + 2ab × a + 2ab × b
= a3 + a2 b + ab2 + b3 + 2a2b + 2ab2
= a3 + b3 + 3a2b + 3ab2
(iii) (a - b)(a - b)(a - b)
In result (ii), replacing b by -b, we get
(a - b)(a - b)(a - b)
= a3 + (-b)3 + 3a2(-b) + 3a(-b)2
= a3 - b3 - 3a2b + 3ab2