Question

factorise (a-b) 3+(b-c) 3+(c-a)3

Answer 1

(a-b)3 + (b-c)3 + (c-a)3 = a3 - b3 - 3ab(a-b) + b3 - c3 - 3bc(b-c) + c3 - a3 - 3ca(c-a)
= - 3a2b + 3ab2 - 3b2c + 3bc2 - 3ac2 + 3 a2c = 3 (- a2b + ab2 - b2c + bc2 - ac2 + a2c)
= 3 [(a2(c-b) + (b2(a-c) + (c2(b-a)]

Or

Let x = (a – b), y = (b – c) and z = (c – a)
Consider, x + y + z = (a – b) + (b – c) + (c – a) = 0
⇒ x3 + y3 + z3 = 3xyz
That is (a – b)3 + (b – c)3 + (c – a)3 = 3(a – b)(b – c)(c – a)



Hope it helped☺☺

Answer 2

Answer:

$(a-b) ^{3} +(b-c)^{3} +(c-a)^{3}$ can be factorized as :

(i) $3(a - b)(b - c)(c - a)$

(ii) $3 [(a^{2} (c-b)) + (b^{2} (a-c)) + (c^{2} (b-a))]$

Explanation:

• Let $x = (a - b), y = (b - c)$ and $z = (c - a)$

• Consider, $x + y + z = (a - b) + (b - c) + (c - a) = 0$
• ⇒ $x^{3} + y^{3} + z^{3} = 3xyz$
• That is, $(a - b)^{3} + (b - c)^{3} + (c - a)^{3} = 3(a -b)(b - c)(c - a)$

Alternatively,

• $(a-b) ^{3} +(b-c)^{3} +(c-a)^{3}$ $= a^{3} - b^{3} - 3ab(a-b) + b^{3} - c^{3} - 3bc(b-c) + c^{3} - a^{3} - 3ca(c-a)$

• $= - 3a^{2} b + 3ab^{2} - 3b^{2} c + 3bc^{2} - 3ac^{2} + 3 a^{2} c = 3 (- a^{2} b + ab^{2} - b^{2} c + bx^{2} - ac^{2} + a^{2} c)$
• $= 3 [(a^{2} (c-b)) + (b^{2} (a-c)) + (c^{2} (b-a))]$