Chapter 6 - Factorisation of Algebraic Expressions
Practice set 6.1
1. Factorise.
(1) x² + 9x + 18
Solution:
x² + 9x + 18
= x² + 6x + 3x + 18
= x (x + 6) + 3(x + 6)
= (x + 6) (x + 3)
∴ x² + 9x + 18 = (x + 6) (x + 3)
(2) x² – 10x + 9
Solution:
x² – 10x + 9
= x² – 9x – x + 9
= x (x – 9) – 1(x – 9)
= (x – 9) (x – 1)
∴ x² – 10x + 9 = (x – 9) (x – 1)
(3) y² + 24y + 144
Solution:
y² + 24y + 144
= y² + 12y + 12y + 144
= y(y + 12) + 12(y + 12)
= (y + 12) (y + 12)
∴ y² + 24y + 144 = (y + 12) (y + 12)
(4) 5y² + 5y – 10
Solution:
5y² + 5y – 10
= 5 (y² + y – 2)
= 5 (y² + 2y – y – 2)
= 5 [y(y + 2) – 1(y + 2)]
= 5 (p + 2) (y – 1)
∴ 5y² + 5y – 10 = 5 (p + 2) (y – 1)
(5) p² – 2p – 35
Solution:
p² – 2p – 35
= p² – 7p + 5p – 35
= p(p – 7) + 5(p – 7)
= (p – 7) (p + 5)
∴ p² – 2p – 35 = (p – 7) (p + 5)
(6) p² – 7p – 44
Solution:
p² – 7p – 44
= p² – 11p + 4p – 44
= p(p – 11) + 4(p – 11)
= (p – 11) (p + 4)
∴ p² – 7p – 44 = (p – 11) (p + 4)
(7) m² – 23m + 120
Solution:
m² – 23m + 120
= m² – 15m – 8m + 120
= m (m – 15) – 8 (m – 15)
= (m – 15) (m – 8)
∴ m² – 23m + 120 = (m – 15) (m – 8)
(8) m² – 25m + 100
Solution:
m² – 25m + 100
= m² – 20m – 5m + 100
= m(m – 20) – 5(m – 20)
= (m – 20) (m – 5)
∴ m² – 25m + 100 = (m – 20) (m – 5)
(9) 3x² + 14x + 15
Solution:
3x² + 14x + 15
= 3x² + 9x + 5x + 15
= 3x(x + 3) + 5(x + 3)
= (x + 3) (3x + 5)
∴ 3x² + 14x + 15 = (x + 3) (3x + 5)
(10) 2x² + x – 45
Solution:
2x² + x – 45
= 2x² + 10x – 9x – 45
= 2x(x + 5) – 9 (x + 5)
= (x + 5) (2x – 9)
∴ 2x² + x – 45 = (x + 5) (2x – 9)
(11) 20x² – 26x + 8
Solution:
20x² – 26x + 8
= 2 (10x² – 13x + 4)
= 2 (10x² – 8x – 5x + 4)
= 2 [2x(5x – 4) – 1(5x – 4)]
= 2 (5x – 4) (2x – 1)
∴ 20x² – 26x + 8 = 2 (5x – 4) (2x – 1)
(12) 44x² – x – 3
Solution:
44x² – x – 3
= 44x² – 12x + 11x – 3
= 4x(11x – 3) + 1(11x – 3)
= (11x – 3) (4x + 1)
∴ 44x² – x – 3 = (11x – 3) (4x + 1)
Practice set 6.2
1. Factorise.
(1) x³ + 64y³
Solution:
x³ + 64y³
= x³ + (4y)³
Here, a = x and b = 4y
We know that,
a³ + b³ = (a + b)(a² – ab + b²)
∴ x³ + 64y³
= (x + 4y) [x² – x(4y) + (4y)²]
= (x + 4y)(x² – 4xy + 16y²)
∴ x³ + 64y³ = (x + 4y)(x² – 4xy + 16y²)
(2) 125p³ + q³
Solution:
125p³ + q³
= (5p)³ + q³
Here, a = 5p and b = q
We know that,
a³ + b³ = (a + b)(a² – ab + b²)
∴ 125p³ + q³
= (5p + q)[(5p)² – (5p)(q) + q²]
= (5p + q)(25p² – 5pq + q²)
∴ 125p³ + q³ = (5p + q)(25p² – 5pq + q²)
(3) 125k³ + 27m³
Solution:
125k³ + 27m³
= (5k)³ + (3m)³
Here, a = 5k and b = 3m
We know that,
a³ + b³ = (a + b)(a² – ab + b²)
∴ 125k³ + 27m³
= (5k + 3m) [(5k)² – (5k)(3m) + (3m)²]
= (5k + 3m)(25k² – 15km + 9m²)
∴ 125k³ + 27m³ = (5k + 3m)(25k² – 15km + 9m²)
(4) 2l³ + 432m³
Solution:
2l³ + 432m³
= 2 (l³ + 216m³) …[Common factor is 2]
= 2 [l³ + (6m)³]
Here, a = x and b = 4y
We know that,
a³ + b³ = (a + b)(a² – ab + b²)
∴ 2l³ + 432m³
= 2 {(l + 6m) [l² – l(6m) + (6m)²]}
= 2 (l + 6m) (l² – 6lm + 36m²)
∴ 2l³ + 432m³ = 2 (l + 6m) (l² – 6lm + 36m²)
(5) 24a³ + 81b³
Solution:
24a³ + 81b³
= 3 (8a³ + 27b³) …[Common factor is 3]
= 3 [(2a)³ + (3b)³]
Here, a = 2a and b = 3b
We know that,
a³ + b³ = (a + b)(a² – ab + b²)
∴ 24a³ + 81b³
= 3 {(2a + 3b) [(2a)² – (2a)(3b) + (3b)²]}
= 3(2a + 3b)(4a² – 6ab + 9b²)
∴ 24a³ + 81b³ = 3(2a + 3b)(4a² – 6ab + 9b²)
(6) y³ + \(\large \frac {1}{8y³}\)
Solution:
y³ + \(\large \frac {1}{8y³}\)
= y³ + \(\large (\frac {1}{2y})\)³
Here, a = y and b = \(\large \frac {1}{2y}\)
We know that,
a³ + b³ = (a + b)(a² – ab + b²)
∴ y³ + \(\large \frac {1}{8y³}\)
= \(\large (\small y + \large (\frac {1}{2y})\)) \(\large [\small y² – y(\frac {1}{2y}) + \large (\frac {1}{2y})²]\)
= \(\large (\small y + \large (\frac {1}{2y})\)) \(\large (\small y² – \frac {1}{2} + \large \frac {1}{4y²})\)
∴ y³ + \(\large \frac {1}{8y³}\) = \(\large (\small y + \large (\frac {1}{2y})\)) \(\large (\small y² – \frac {1}{2} + \large \frac {1}{4y²})\)
(7) a³ + \(\large \frac {8}{a³}\)
Solution:
a³ + \(\large \frac {8}{a³}\)
= a³ + \(\large (\frac {2}{a})\)³
Here, a = a and b = \(\large \frac {2}{a}\)
We know that,
a³ + b³ = (a + b)(a² – ab + b²)
∴ a³ + \(\large \frac {8}{a³}\)
= \(\large (\small a + \large (\frac {2}{a})\)) \(\large [\small a² – a(\frac {2}{a}) + \large (\frac {2}{a})²]\)
= \(\large (\small a + \large (\frac {2}{a})\)) \(\large (\small a² – 2 + \large \frac {4}{a²})\)
∴ a³ + \(\large \frac {8}{a³}\) = \(\large (\small a + \large (\frac {2}{a})\)) \(\large (\small a² – 2 + \large \frac {4}{a²})\)
(8) 1 + \(\large \frac {q³}{125}\)
Solution:
1 + \(\large \frac {q³}{125}\)
= 1³ + \(\large (\frac {q}{5})\)³
Here, a = 1 and b = \(\large \frac {q}{5}\)
We know that,
a³ + b³ = (a + b)(a² – ab + b²)
∴ 1 + \(\large \frac {q³}{125}\)
= \(\large (\small 1 + \large (\frac {q}{5})\)) \(\large [\small 1² – 1(\frac {q}{5}) + \large (\frac {q}{5})²]\)
= \(\large (\small 1 + \large (\frac {q}{5})\)) \(\large (\small 1 – \large (\frac {q}{5}) + \large \frac {q²}{24})\)
∴ 1 + \(\large \frac {q³}{125}\) = \(\large (\small 1 + \large (\frac {q}{5})\)) \(\large (\small 1 – \large (\frac {q}{5}) + \large \frac {q²}{24})\)
Practice set 6.3
1. Factorise :
(1) y³ – 27
Solution:
y³ – 27
= y³ – (3)³
Here, a = y and b = 3
We know that,
a³ – b³ = (a – b) (a² + ab + b²)
∴ y³ – 27
= (y – 3)[y² + y(3) + (3)2]
= (y – 3)(y² + 3y + 9)
∴ y³ – 27 = (y – 3)(y² + 3y + 9)
(2) x³ – 64y³
Solution:
x³ – 64y³
= x³ – (4y)³
Here, a = x and b = 4y
We know that,
a³ – b³ = (a – b) (a² + ab + b²)
∴ x³ – 64y³
= (x – 4y)[x² + x(4y) + (4y)²]
= (x – 4y)(x² + 4xy + 16y²)
∴ x³ – 64y³ = (x – 4y)(x² + 4xy + 16y²)
(3) 27m³ – 216n³
Solution:
27m³ – 216n³
= 27 (m³ – 8n³)
= 27 [m³ – (2n)³]
Here, a = m and b = 2n
We know that,
a³ – b³ = (a – b) (a² + ab + b²)
∴ 27m³ – 216n³
= 27 {(m – 2n) [m² + m(2n) + (2n)²]}
= 27 (m – 2n)(m² + 2mn + 4n²)
∴ 27m³ – 216n³ = 27 (m – 2n)(m² + 2mn + 4n²)
(4) 125y³ – 1
Solution:
125y³ – 1
= (5y)³ – 1³
Here, a = 5y and b = 1
We know that,
a³ – b³ = (a – b) (a² + ab + b²)
∴ 125y³ – 1
= (5y – 1) [(5y)² + (5y)(1) + (1)²]
= (5y – 1) (25y² + 5y + 1)
∴ 125y³ – 1 = (5y – 1) (25y² + 5y + 1)
(5) 8p³ – \(\large \frac {27}{p³}\)
Solution:
8p³ − \(\large \frac {27}{p³}\)
= (2p)³ – \(\large (\frac {3}{p})³\)
Here,
a = 2p and b = \(\large \frac {3}{p}\)
We know that,
a³ – b³ = (a – b) (a² + ab + b²)
∴ 8p³ − \(\large \frac {27}{p³}\)
= \(\large(\)2p – \(\large (\frac {3}{p})\) \(\large[\)(2p)² + \(\large (2p) (\frac {3}{p})\) + \(\large \frac {3}{p}]\)²
= \(\large(\)2p – \(\large (\frac {3}{p})\) \(\large[\)4p² + 6 + \(\large \frac {9}{p²}]\)
∴ 8p³ − \(\large \frac {27}{p³}\) = \(\large(\)2p – \(\large (\frac {3}{p})\) \(\large[\)4p² + 6 + \(\large \frac {9}{p²}]\)
(6) 343a³ – 512b³
Solution:
343a³ – 512b³
= (7a)³ – (8b)³
Here, A = 7a and B = 8b
We know that,
a³ – b³ = (a – b) (a² + ab + b²)
∴ 343a³ – 512b³
= (7a – 8b) [(7a)² + (7a)(8b) + (8b)²]
= (7a – 8b) (49a² + 56ab + 64b²)
∴ 343a³ – 512b³ = (7a – 8b) (49a² + 56ab + 64b²)
(7) 64x³ – 729y³
Solution:
64x³ – 729y³
= (4x)³ – (9y)³
Here, a = 4x and b = 9y
We know that,
a³ – b³ = (a – b) (a² + ab + b²)
∴ 64x³ – 729y³
= (4x – 9y) [(4x)² + (4x) (9y) + (9y)²]
= (4x – 9y) (16x² + 36xy + 81y²)
∴ 64x³ – 729y³ = (4x – 9y) (16x² + 36xy + 81y²)
(8) 16a³ − \(\large \frac {128}{b³}\)
Solution:
16a³ − \(\large \frac {128}{b³}\)
= 16 \(\large[\)a³ – \(\large \frac {8}{b³}]\)
= 16 \(\large[\)a³ – \(\large (\frac {2}{b})³]\)
Here,
A = a and B = \(\large \frac {2}{b}\)
We know that,
a³ – b³ = (a – b) (a² + ab + b²)
∴ 16a³ − \(\large \frac {128}{b³}\)
= 16 \(\large(\)a – \(\large (\frac {2}{b})\) \(\large[\)a² + \(\large a(\frac {2}{b})\) + \(\large \frac {2}{b}]\)²
= 16 \(\large(\)a – \(\large (\frac {2}{b})\) \(\large[\)a² + \(\large \frac {2a}{b}\) + \(\large \frac {4}{b²}\)
∴ 16a³ − \(\large \frac {128}{b³}\) = 16 \(\large(\)a – \(\large (\frac {2}{b})\) \(\large[\)a² + \(\large \frac {2a}{b}\) + \(\large \frac {4}{b²}\)
2. Simplify :
(1) (x + y)³ – (x – y)³
Solution:
Here, a = x + y and b = x – y
We know that,
a³ – b³ = (a – b)(a² + ab + b²)
∴ (x + y)³ – (x – y)³
= [(x + y) – (x – y)] [(x + y)² + (x + y) (x – y) + (x – y)]
= (x + y – x + y) [(x² + 2xy + y²) + (x² – y²) + (x² – 2xy + y²)]
= 2y(x² + x² + x² + 2xy – 2xy + y² – y² + y²)
= 2y (3x² + y²)
= 6x²y + 2y³
∴ (x + y)³ – (x – y)³ = 6x²y + 2y³
(2) (3a + 5b)³ – (3a – 5b)³
Solution:
Here, A = 3a + 5b and B = 3a – 5b
We know that,
A³ – B³ = (A – B)(A² + AB + B²)
∴ (3a + 5b)³ – (3a – 5b)³
= [(3a + 5b) – (3a – 5b)] [(3a + 5b)² + (3a + 5b) (3a – 5b) + (3a – 5b)²]
= (3a + 5b – 3a + 5b) [(9a² + 30ab + 25b²) + (9a² – 25b²) + (9a² – 30ab + 25b²)]
= 10b (9a² + 9a² + 9a² + 30ab – 30ab + 25b² – 25b² + 25b²)
= 10b (27a² + 25b²)
= 270a²b + 250b³
∴ (3a + 5b)³ – (3a – 5b)³ = 270a²b + 250b³
(3) (a + b)³ – a³ – b³
Solution:
We know that,
a³ – b³ = (a – b)(a² + ab + b²)]
∴ (a + b)³ – a³ – b³
= a³ + 3a²b + 3ab² + b³ – a³ – b³
= 3a²b + 3ab²
∴ (a + b)³ – a³ – b³ = 3a²b + 3ab²
(4) p³ – (p + 1)³
Solution:
We know that,
a³ – b³ = (a – b)(a² + ab + b²)]
∴ p³ – (p + 1)³
= p³ – (p³ + 3p² + 3p + 1) …[∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
= p³ – p³ – 3p² – 3p – 1
= – 3p² – 3p – 1
∴ p³ – (p + 1)³ = – 3p² – 3p – 1
(5) (3xy – 2ab)³ – (3xy + 2ab)³
Solution:
Here, A = 3xy – 2ab and B = 3xy + 2ab
We know that,
A³ – B³ = (A – B)(A² + AB + B²)
∴ (3xy – 2ab)³ – (3xy + 2ab)³
= [(3xy – 2ab) – (3xy + 2ab)] [(3xy – 2ab)² + (3xy – 2ab) (3xy + 2ab) + (3xy + 2ab)²]
= (3xy – 2ab – 3xy – 2ab) [(9x²y² – 12xyab + 4a²b²) + (9x²y² – 4a²b²) + (9x²y² + 12xyab + 4a²b²)]
= (– 4ab) (9x²y² + 9x²y² + 9x²y² – 12xyab + 12xyab + 4a²b² – 4a²b² + 4a²b²)
= (– 4ab) (27 xy² + 4a²b²)
= –108x²y²ab – 16a³b³
∴ (3xy – 2ab)³ – (3xy + 2ab)³ = –108x²y²ab – 16a³b³
Practice set 6.4
1. Simplify
1. \(\large \frac {m²\,–\,n²}{(m\,+\,n)²}\) × \(\large \frac {m²\,+\,mn\,+\,n²}{m³\,–\,n³}\)
Solution:
\(\large \frac {m²\,–\,n²}{(m\,+\,n)²}\) × \(\large \frac {m²\,+\,mn\,+\,n²}{m³\,–\,n³}\)
= \(\large \frac {(m\,+\,n)(m\,–\,n)}{(m\,+\,n)(m\,+\,n)}\) × \(\large \frac {m²\,+\,mn\,+\,n²}{(m\,–\,n)(m²\,+\,mn\,+\,n²)}\)
= \(\large \frac {1}{(m\,+\,n}\)
∴ \(\large \frac {m²\,–\,n²}{(m\,+\,n)²}\) × \(\large \frac {m²\,+\,mn\,+\,n²}{m³\,–\,n³}\) = \(\large \frac {1}{(m\,+\,n}\)
2. \(\large \frac {a²\,+\,10a\,+\,21}{a²\,+\,6a\,–\,7}\) × \(\large \frac {a²\,–\,1}{a\,+\,3}\)
Solution:
\(\large \frac {a²\,+\,10a\,+\,21}{a²\,+\,7a\,–\,a–\,7}\) × \(\large \frac {a²\,–\,1}{a\,+\,3}\)
= \(\large \frac {a²\,+\,7a\,+\,3a\,+\,21}{a²\,+\,6a\,–\,7}\) × \(\large \frac {a²\,–\,1²}{a\,+\,3}\)
= \(\large \frac {a(a\,+\,7)\,+\,3(a\,+\,7)}{a(a\,+\,7)\,–\,1(a\,+\,7)}\) × \(\large \frac {(a\,+\,1)(a\,–\,1)}{a\,+\,3}\)
= a + 1
∴ \(\large \frac {a²\,+\,10a\,+\,21}{a²\,+\,6a\,–\,7}\) × \(\large \frac {a²\,–\,1}{a\,+\,3}\) = a + 1
3. \(\large \frac {8x³\,–\,27y³}{4x²\,–\,9y²}\)
Solution:
\(\large \frac {8x³\,–\,27y³}{4x²\,–\,9y²}\)
= \(\large \frac {(2x)³\,–\,(3y)³}{(2x)²\,–\,(3y)²}\)
= \(\large \frac {(2x\,–\,3y)[(2x)²\,+\,(2x)(3y)\,+\,(3y)²}{(2x\,+\,3y)(2x\,–\,3y)}\)
= \(\large \frac {4x²\,+\,6xy\,+\,9y²}{2x\,+\,3y}\)
∴ \(\large \frac {8x³\,–\,27y³}{4x²\,–\,9y²}\) = \(\large \frac {4x²\,+\,6xy\,+\,9y²}{2x\,+\,3y}\)
4. \(\large \frac {x²\,–\,5x\,–\,24}{(x\,+\,3)(x\,+\,8)}\) × \(\large \frac {x²\,–\,64}{(x\,–\,8)²}\)
Solution:
\(\large \frac {x²\,–\,5x\,–\,24}{(x\,+\,3)(x\,+\,8)}\) × \(\large \frac {x²\,–\,64}{(x\,–\,8)²}\)
= \(\large \frac {x²\,–\,8x\,+\,3x\,–\,24}{(x\,+\,3)(x\,+\,8)}\) × \(\large \frac {(x)²\,–\,(8)²}{(x\,–\,8)²}\)
= \(\large \frac {x(x\,–\,8)\,+\,3(x\,–\,8)}{(x\,+\,3)(x\,+\,8)}\) × \(\large \frac {(x\,+\,8)(x\,–\,8)}{(x\,–\,8)(x\,–\,8)}\)
= \(\large \frac {(x\,–\,8)(x\,+\,3)}{(x\,+\,3)(x\,+\,8)}\) × \(\large \frac {(x\,+\,8)(x\,–\,8)}{(x\,–\,8)(x\,–\,8)}\)
= 1
∴ \(\large \frac {x²\,–\,5x\,–\,24}{(x\,+\,3)(x\,+\,8)}\) × \(\large \frac {x²\,–\,64}{(x\,–\,8)²}\) = 1
5. \(\large \frac {3x²\,–\,x\,–\,2}{x²\,–\,7x\,+\,12}\) ÷ \(\large \frac {3x²\,–\,7x\,–\,6}{x²\,–\,4}\)
Solution:
\(\large \frac {3x²\,–\,x\,–\,2}{x²\,–\,7x\,+\,12}\) ÷ \(\large \frac {3x²\,–\,7x\,–\,6}{x²\,–\,4}\)
= \(\large \frac {3x²\,–\,x\,–\,2}{x²\,–\,7x\,+\,12}\) × \(\large \frac {x²\,–\,4}{3x²\,–\,7x\,–\,6}\)
= \(\large \frac {3x²\,–\,3x\,+\,2x\,–\,2}{x²\,–\,4x\,–\,3x\,+\,12}\) × \(\large \frac {(x)²\,–\,(2)²}{3x²\,–\,9x\,+\,2x\,–\,6}\)
= \(\large \frac {3x(x\,–\,1)\,+\,2(x\,–\,1)}{x(x\,–\,4)\,–\,3(x\,–\,4)}\) × \(\large \frac {(x\,+\,2)(x\,–\,2)}{3x(x\,–\,3)\,+\,(x\,–\,3)}\)
= \(\large \frac {(x\,–\,1)(3x\,+\,2)}{(x\,–\,4)(x\,–\,3)}\) × \(\large \frac {(x\,+\,2)(x\,–\,2)}{(x\,–\,3)(3x\,+\,2)}\)
= \(\large \frac {(x\,–\,1)(x\,+\,2)(x\,–\,2)}{(x\,–\,4)(x\,–\,3)²}\)
∴ \(\large \frac {3x²\,–\,x\,–\,2}{x²\,–\,7x\,+\,12}\) ÷ \(\large \frac {3x²\,–\,7x\,–\,6}{x²\,–\,4}\) = \(\large \frac {(x\,–\,1)(x\,+\,2)(x\,–\,2)}{(x\,–\,4)(x\,–\,3)²}\)
6. \(\large \frac {4x²\,–\,11x\,+\,6}{16x²\,–\,9}\)
Solution:
\(\large \frac {4x²\,–\,11x\,+\,6}{16x²\,–\,9}\)
= \(\large \frac {4x²\,–\,8x\,–\,3x\,+\,6}{(4x)²\,–\,(3)²}\)
= \(\large \frac {4x(x\,–\,2)\,–\,3(x\,–\,2)}{(4x\,–\,3)(4x\,–\,3)}\)
= \(\large \frac {(x\,–\,2)(4x\,–\,3)}{(4x\,–\,3)(4x\,–\,3)}\)
= \(\large \frac {x\,–\,2}{4x\,–\,3}\)
∴ \(\large \frac {4x²\,–\,11x\,+\,6}{16x²\,–\,9}\) = \(\large \frac {x\,–\,2}{4x\,–\,3}\)
7. \(\large \frac {a³\,–\,27}{5a²\,–\,16a\,+\,3}\) ÷ \(\large \frac {a²\,+\,3a\,+\,9}{25a²\,–\,1}\)
Solution:
\(\large \frac {a³\,–\,27}{5a²\,–\,16a\,+\,3}\) ÷ \(\large \frac {a²\,+\,3a\,+\,9}{25a²\,–\,1}\)
= \(\large \frac {a³\,–\,27}{5a²\,–\,16a\,+\,3}\) × \(\large \frac {25a²\,–\,1}{a²\,+\,3a\,+\,9}\
= \(\large \frac {(a\,–\,3)[(a)²\,+\,(a)(3)\,+\,(3)²]}{5a(a\,–\,3)\,–\,1(a\,–\,3)}\) × \(\large \frac {(5a\,+\,1)(5a\,–\,1)}{a²\,+\,3a\,+\,9}\
= \(\large \frac {(a\,–\,3)(a²\,+\,3a\,+\,9)}{(a\,–\,3)(5a\,–\1)}\) × \(\large \frac {(5a\,+\,1)(5a\,–\,1)}{a²\,+\,3a\,+\,9}\
= 5a + 1
∴ \(\large \frac {a³\,–\,27}{5a²\,–\,16a\,+\,3}\) ÷ \(\large \frac {a²\,+\,3a\,+\,9}{25a²\,–\,1}\) = 5a + 1
8. \(\large \frac {1\,–\,2x\,+\,x²}{1\,–\,x³}\) × \(\large \frac {1\,+\,x\,+\,x²}{1\,+\,x}\)
Solution:
\(\large \frac {1\,–\,2x\,+\,x²}{1\,–\,x³}\) × \(\large \frac {1\,+\,x\,+\,x²}{1\,+\,x}\)
= \(\large \frac {1\,–\,x\,–\,x\,+\,x²}{(1)³\,–\,(x)³}\) × \(\large \frac {1\,+\,x\,+\,x²}{1\,+\,x}\)
= \(\large \frac {1(1\,–\,x)\,–\,x(1\,–\,x)}{(1\,–\,x)[(1)²\,+\,(1)(x)\,+\,(x)²}\) × \(\large \frac {1\,+\,x\,+\,x²}{1\,+\,x}\)
= \(\large \frac {(1\,–\,x)(1\,–\,x)}{(1\,–\,x)(1\,+\,x\,+\,x²}\) × \(\large \frac {1\,+\,x\,+\,x²}{1\,+\,x}\)
= \(\large \frac {1\,–\,x}{1\,+\,x}\)
∴ \(\large \frac {1\,–\,2x\,+\,x²}{1\,–\,x³}\) × \(\large \frac {1\,+\,x\,+\,x²}{1\,+\,x}\) = \(\large \frac {1\,–\,x}{1\,+\,x}\)