Chapter 5: Factorization of Algebraic Expressions Exercise – 5.4
Question: 1
Find the value
a3 + 8b3 + 64c3 − 24abc
Solution:
= (a)3 + (2b)3 + (4c)3 − 3 × a × 2b × 4c
= (a + 2b + 4c)(a2 + (2b)2 + (4c)2 − a × 2b − 2b × 4c − 4c × a)
[∴ a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)]
= (a + 2b + 4c)(a2 + 4b2 + 16c2 − 2ab − 8bc − 4ac)
∴ a3 + 8b3 + 64c3 − 24abc
= (a + 2b + 4c)(a2 + 4b2 + 16c2 − 2ab − 8bc − 4ac)
Question: 2
Find the value
x3 − 8y3 + 27z3 + 18xyz
Solution:
= x3 − (2y)3 + (3z)3 − 3 × x × (−2y)(3z)
= (x + (−2y) + 3z)(x2 + (−2y)2 + (3z)2 − x(−2y) − (−2y)(3z) − 3z(x))
[∴ a3 + b3 + c3 − 3abc = (a + b + c)(a2 + b2 + c2 − ab − bc − ca)]
= (x + (−2y) + 3z)(x2 + 4y2 + 9z2 + 2xy + 6yz − 3zx)
∴ x3 − 8y3 + 27z3 + 18xyz
= (x + (−2y) + 3z)(x2 + 4y2 + 9z2 + 2xy + 6yz − 3zx)
Question: 3
Find the value
Solution:
= (x/3)3 + (−y)3 + (5z)3 − 3 × x/3(−y)(5z)
= (x/3 + (−y) + 5z)((x/3)2 + (−y)2 + (5z)2 − x/3(−y) − (−y)5z − 5z(x/3))
= (x/3 − y + 5z)(x2/9 + y2 + 25z2 + xy/3 + 5yz − 5zx/3)
= (x/3 − y + 5z)(x2/9 + y2 + 25z2 + xy/3 + 5yz − 5zx/3)
Question: 4
Find the value
8x3 + 27y3 − 216z3 + 108xyz
Solution:
= (2x)3 + (3y)3 + (− 6y)3 − 3(2x)(3y)(−6z)
= (2x + 3y + (− 6z))((2x)2 + (3y)2 + (−6z)2 − 2x × 3y − 3y(−6z) − (−6z)2x)
= (2x + 3y + (− 6z))(4x2 + 9y2 + 36z2 − 6xy + 18yz + 12zx)
? 8x3 + 27y3 − 216z3 + 108xyz
= (2x + 3y + (− 6z))(4x2 + 9y2 + 36z2 − 6xy + 18yz + 12zx)
Question: 5
Find the value
125 + 8x3 − 27y3 + 90xy
Solution:
= (5)3 + (2x)3 + (−3y)3 − 3 × 5 × 2x × (−3y)
= (5 + 2x + (−3y))(52 + (2x)2 + (−3y)2 − 5(2x) − 2x(−3y) − (−3y)5)
= (5 + 2x − 3y)(25 + 4x2 + 9y2 − 10x + 6xy + 15y)
∴ 125 + 8x3 − 27y3 + 90xy
= (5 + 2x − 3y)(25 + 4x2 + 9y2 − 10x + 6xy + 15y)
Question: 6
Find the value
(3x − 2y)3 + (2y − 4z)3 + (4z − 3x)3
Solution:
Let (3x − 2y) = a, (2y − 4z) = b, (4z − 3x) = c
∴ a + b + c = 3x − 2y + 2y − 4z + 4z − 3x = 0
∵ a + b + c = 0 ∴ a3 + b3 + c3 = 3abc
= 3(3x − 2y)(2y − 4z)(4z − 3x)
∴ (3x − 2y)3 + (2y − 4z)3 + (4z − 3x)3
= 3(3x − 2y)(2y − 4z)(4z − 3x)
Question: 7
Find the value
(2x − 3y)3 + (4z − 2x)3 + (3y − 4z)3
Solution:
Let 2x - 3y = a, 4z - 2x = b, 3y - 4z = c
∴ a + b + c = 2x − 3y + 4z − 2x + 3y − 4z = 0
∵ a + b + c = 0 ∴ a3 + b3 + c3 = 3abc
= 3(2x − 3y)(4z − 2x)(3y − 4z)
∴ (2x − 3y)3 + (4z − 2x)3 + (3y − 4z)3
= 3(2x − 3y)(4z − 2x)(3y − 4z)
Question: 8
Find the value
[x/2 + y + z/3]3 + [x/3 − 2y/3 + z]3 + [−5x/6 − y/3 − 4z/3]3
Solution:
Let [x/2 + y + z/3] = a,[x/3 − 2y/3 + z] = b, [−5x/6 − y/3 − 4z/3] = c
a + b + c = x/2 + y + z/3 + x/3 − 2y/3 + z − 5x/6 − y/3 − 4z/3
a + b + c = (x/2 + x/3 − 5x/6) + (y − 2y/3 − y/3) + (z/3 + z − 4z/3)
a + b + c = 3x/6 + 2x/6 − 5x/6 + 3y/3 − 2y/3 − y/3 + z/3 + 3z/3 − 4z/3
a + b + c = 0
∵ a + b + c = 0 ∴ a3 + b3 + c3 = 3abc
= 3(x/2 + y + z/3)(x/3 − 2y/3 + z)(−5x/6 − y/3 − 4z/3)
∴ [x2 + y + z3]3 + [x3 − 2y3 + z]3 + [−5x/6 − y/3 − 4z/3]3
= 3(x/2 + y + z/3)(x/3 − 2y/3 + z)(−5x/6 − y/3 − 4z/3)
Question: 9
Find the value
(a − 3b)3 + (3b − c)3 + (c − a)3
Solution:
Let a - 3b = x, 3b - c = y, c - a = z
x + y + z = a − 3b + 3b − c + c − a = 0
(∴ x + y + z = 0)?
x3 + y3 + z3 = 3xyz
= 3(a − 3b)(3b − c)(c − a)
∴ (a − 3b)3 + (3b − c)3 + (c − a)3
= 3(a − 3b)(3b − c)(c − a)
Question: 10
Find the value
Solution:
Question: 11
Find the value
Solution:
Question: 12
Find the value
8x3 − 125y3 + 216 + 180xy
Solution:
= (2x)3 + (−5y)3 + 63 − 3 × (2x)(−5y)(6)
= (2x + (−5y) + 6)((2x)2 + (−5y)2 + 62 − 2x × (−5y) − (−5y)6 − 6(2x))
= (2x − 5y + 6)(4x2 + 25y2 + 36 + 10xy + 30y − 12x)
∴ 8x3 − 125y3 + 216 + 180xy
= (2x − 5y + 6)(4x2 + 25y2 + 36 + 10xy + 30y − 12x)
Question: 13
Find the value
Solution:
Question: 14
Find the value of x3 + y3 − 12xy + 64 when x + y = - 4.
Solution:
= x3 + y3 + 64 − 12xy
= x3 + y3 + 43 − 3(x)(y)(4)
= (x + y + 4)(x2 + y2 + 42 − xy − y × 4 − 4 × x)
= (−4 + 4)(x2 + y2 + 16 − xy − 4y − 4x)
[∴ x + y = −4] = 0
∴ x3 + y3 − 12xy + 64 = 0
Question: 15
Multiply:
(i) x2 + y2 + z2 − xy + xz + yz by x + y − z
(ii) x2 + 4y2 + z2 + 2xy + xz − 2yzbyx − 2y − z
(iii) x2 + 4y2 + 2xy−3x+6y+9 by (x−2y+3)
(iv) 9x2 + 25y2 + 15xy + 12x − 20y + 16by 3x − 5y + 4
Solution:
(i) x2 + y2 + z2 − xy + xz + yz by x + y − z
= (x2 + y2 + z2 − xy + xz + yz)(x + y − z)
= x3 + y3 + z3 − 3xyz
(ii) x2 + 4y2 + z2 + 2xy + xz − 2yzbyx − 2y − z
x2 + (−2y)2 + (-z)2 − (−2y)(−z) − (−z)(x)
= x3 + (−2y)3 + (-z)3 − 3x(−2y)(−z)
⇒ x2 + 4y2 + z2 + 2xy − 2yz + zx
= x3 − 8y3 − z3 − 6xyz
(iii) x2 + 4y2 + 2xy−3x + 6y + 9 by (x − 2y + 3)
(x)2 + (−2y)2 + (3)2 − (x)(−2y) − (−2y)(3) − 3(x)
= (x)3 + (−2y)3 + 33 − 3(x)(−2y)(3)
⇒ x2 + 4y2 + 9 + 2xy + 6y − 3x
= x3 − 8y3 + 27 + 18xy
(iv) 9x2 + 25y2 + 15xy + 12x − 20y + 16by 3x − 5y + 4
(3x)2 + (5y)2 + 42 − (−3x)(5y) − (5y)(4) − (4)(−3x)
= (−3x)3 + (5y)3 + 43 − 3(−3x)(5y)(4)
⇒ 9x2 + 25y2 + 16 + 15xy − 20y + 12x
= −27x3 + 125y3 + 64 + 180xy