# How to show using factorTheorem,show that a-b,b-c and c-a are the factors of a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2)?

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You need to notice that the differences of squares may be converted into products such that:

`a(b -c)(b + c) + b(c - a)(c + a) + c(a - b)(a + b)`

Notice that the factor theorem states that (`x - a` ) is a factor of a polynomial f(x) if and only if `f(a) = 0` and since the problem does not provide a polynomial, the factor theorem is useless in this case.

Instead, you may openthe brackets such that:

`ab^2 - ac^2 + bc^2 - ba^2 + ca^2 - cb^2`

You may group the terms such that:

`(ab^2 - ba^2) + (ca^2 - ac^2) + (bc^2 - cb^2)`

Factoring out `ab` ,`ac` and `bc` yields:

`ab(a - b) + ac(a - c) + bc(b - c)`

**Notice the presence of partial factors a - b , a - c , b - c, but they are not the factors of entire given expression `a(b^2-c^2)+b(c^2-a^2)+c(a^2-b^2).` **

Here,

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

=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2

=ab^2-ac^2+bc^2-ba^2+ca^2-cb^2+abc-abc

=abc-ba^2-ac^2+ca^2-cb^2+ab^2+bc^2-abc

=ab(c-a)-ac(c-a)-b^2(c-a)+bc(c-a)

=(ab-ac-b^2+bc)(c-a)

={a(b-c)-b(b-c)}(c-a)

=(a-b)(b-c)(c-a)

Therefore (a-b) ,(b-c) and (c-a) are the factors of the given expression.[I see no way to solve this using factor theorem. You get the idea to solve by this method if you multiply the three factors first before solving in this way and thus get the required expression.]