(a-b)^3+(b-c)^3+(c-a)^3
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(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☺☺
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