1. ## Factoring

I have another factoring problem from Gelfand's Algebra book that I've been unable to solve;

Problem 122. Factor:
(f) $(a-b)^3+(b-c)^3+(c-a)^3$

My try:
$(a-b)^3+(b-c)^3+(c-a)^3=((a-b)+(b-c))((a-b)^2-(a-b)(b-c)+(b-c)^2)+(c-a)^3=(a-c)(a^2-2ab+b^2-(ab-ac-b^2+bc)+b^2-2bc-c^2)+(c-a)^3=(a-c)(a^2-2ab+b^2-ab+ac+b^2-bc+b^2-2bc+c^2)-(a-c)^3=(a-c)(a^2-3ab+3b^2+ac-3bc+c^2)-(a-c)^3=?$

2. ## Re: Factoring

Another factoring problem

shows that $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz -zx)$

so if $x+y+z = 0$, then $x^3 + y^3 + z^3 = 3xyz$.

Maybe you can apply this result to the current problem.

3. ## Re: Factoring

Originally Posted by awkward
Another factoring problem

shows that $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2 + y^2 + z^2 - xy - yz -zx)$

so if $x+y+z = 0$, then $x^3 + y^3 + z^3 = 3xyz$.

Maybe you can apply this result to the current problem.
So I guess the answer is $3(a-b)(b-c)(c-a)$?

But isn't $x^3+y^3+z^3$ equal to $3xyz$ only if $x+y+z=0 ?$ or am I asking a stupid question. I mean by letting $x+y+z=0$, $x^3+y^3+z^3$ has to be $3xyz$ but does that mean that $x^3+y^3+z^3$ is generally equal to $3xyz$.

4. ## Re: Factoring

Originally Posted by DIOGYK
So I guess the answer is $3(a-b)(b-c)(c-a)$?

But isn't $x^3+y^3+z^3$ equal to $3xyz$ only if $x+y+z=0 ?$ or am I asking a stupid question. I mean by letting $x+y+z=0$, $x^3+y^3+z^3$ has to be $3xyz$ but does that mean that $x^3+y^3+z^3$ is generally equal to $3xyz$.
Simplify (a-b) + (b-c) + (c-a). What do you get?

5. ## Re: Factoring

Originally Posted by awkward
Simplify (a-b) + (b-c) + (c-a). What do you get?
0 of course, now I get it. I attacked this problem by fists, filled 2-3 papers haha, and you attacked it with your brain, you're awesome. Thanks for help awkward.