How to get -(a-b)(b-c)(c-a) from ab(a-b)+bc(b-c)+ca(c-a)?

Question

I was wondering if anyone could really explain this to me. I came across a formula which states : ab(a-b)+bc(b-c)+ca(c-a) = -(a-b)(b-c)(c-a)

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I am able to work the problem our from -(a-b)(b-c)(c-a) to get ab(a-b)+bc(b-c)+ca(c-a); however, I am not able to get it the other way around.

E.g.,: -(a-b)(b-c)(c-a)

-(ab - ac - b² + bc)(c-a)

-(abc - a²b - ac² + a²c - b²c + ab² + bc² - abc)

- abc + a²b + ac² - a²c + b²c - ab² - bc² + abc

a²b + ac² - a²c + b²c - ab² - bc²

Rearrange:

(a²b - ab²) + (b²c - bc²) + (ac�� - a²c)

ab(a-b)+bc(b-c)+ca(c-a)

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So, I was just a bit curious about it.

So I was mainly wondering how to get -(a-b)(b-c)(c-a) from ab(a-b)+bc(b-c)+ca(c-a)

Sepia · Accepted Answer

ab(a - b) + bc(b - c) + ca(c - a)
= a²b - ab² + b²c - bc² + c²a - ca²
= a²b - a²c - ab² + ac² + b²c - bc²
= a²(b - c) - a(b² - c²) + bc(b - c)
= a²(b - c) - a(b + c)(b - c) + bc(b - c)
= (b - c)[a² - a(b + c) + bc]
= (b - c)(a - b)(a - c)(b - c)
= (a - b)(a - c)(b - c)
= -(a - b)(b - c)(c - a)

? · Answer

You would have to know to add an abc then subtract the same thing which would cancel,  but is necessary to make the factors.

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How to get -(a-b)(b-c)(c-a) from ab(a-b)+bc(b-c)+ca(c-a)?

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