Question

show that (a-b)3+(b-c)3+(c-a)3=3(a-b)(b-c)(c-a)​

Answer 1

Step-by-step explanation:

We know, (a−b)

3 =a 3 −b 3 −3a 2 b+3ab 2

⟹ (a−b) 3

=a 3 −b 3 −3ab(a−b).

Then,

(a−b) 3 +(b−c) 3 +(c−a) 3

=(a 3 −3a 2 b+3ab 2 −b 3 )+(b 3 −3b 2 c+3bc 2 −c 3 )+(c 3 −3c 2 a+3ca 2 −a 3 )

=−3a 2 b+3ab 2 −3b 2 c+3bc 2 −3c 2 a+3ca 2

=3a 2 (c−b)+3b 2 (a−c)+3c 2 (b−a).

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Answer 2

Answer:

(a-b) (b-c)^2 (c-a)^2 (c-b) (9)

Step-by-step explanation:

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

(a-b) [+(b-c) (b-c)] [+(c-a) (c-a)] (c-b) (3+3+3)

(a-b) (b-c)^2 (c-a)^2 (c-b)(9)

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