Question

(a-b)^3 - (b-c)^3 +(c-a)^3 + 3(a-b)(b-c)(c-a)​

Answer 1

Answer:

This is a cyclic expression of degree 3. From very old high school classes, that is enough to deduce the factors. First, notice that a=b , a=c and b=c , each cause a value of 0 . So each of a-b, b-c and c-a are factors and (a-b)(b-c)(c-a) produces a cyclic expression of degree 3. Therefore.

(a−b)3+(b−c)3+(c−a)3=K(a−b)(b−c)(c−a)

Answer 2

Step-by-step explanation:

Consider (a-b)^3 + (b-c)^3 + (c-a)^3 - 3(a-b)(b-c)(c-a).

We want to show that this is identically equal to zero. It is clearly zero if a = b = c. Differentiating partially with respect to a,

3(a-b)^2 - 3(c-a)^2 - 3(b-c)(c-a) + 3(a-b)(b-c) = 3(a-b)(a-c) - 3(c-a)(b-a) = 0.

Therefore, for any values of b and c, the function is a constant with respect to a. Similarly it is constant as a function of b or of c.

This proves the result.

I have used factoring which is complementary to expanding. I hope that is within your arbitrary rules.

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