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the expression is $(a-b)c^3+(b-c)a^3-(a-c)b^3$

I've tried factoration by grouping, but it has been a trial and error situation. I don't know any strategies to factor expressions like that, so I'm basically just guessing.

There is a ton of questions like this one in the book I'm studying right now and I can't solve anything, it just feels like I have to guess since I don't know any method to do this.

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    It is written exactly like that. No typos. – 2017-01-12

2 Answers 2

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One strategy proceeds as follows. If $a=c$ then the expression becomes $(a-b)a^3+(b-a)a^3$ which is equal to zero so $(a-c)$ is a factor. This is a simple extension of the remainder theorem for polynomials - if we have a polynomial $p(x)$ and we find that $p(a)=0$ then $(x-a)$ is a factor.

You should be able to show also that $(b-c)$ and $(a-b)$ are factors. Then the whole expression you have has degree $4$, so there is only one factor left. To get terms involving $a^3, b^3, c^3$ there are not many options for the final factor.

So this involves some guessing, but simple guessing. For similar expressions it is useful to become adept at working with symmetric polynomials.

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    Aw dang it, this is what I get for trying to work the entire problem out: I answer slower. :D – 2017-01-12
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    @SimpleArt There are people who have done that to me many times ... – 2017-01-12
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One may first set the expression equal to $0$.

$$0=(a-b)c^3+(b-c)a^3+(c-a)b^3$$

One can then see that for this to hold, we have one solution $a=b$, $a=c$, and $b=c$. Turning this into "factors" that we can use, we get, as a polynomial of $a$,

$$(a-b)(a-c)(b-c)P(a)=(a-b)c^3+(b-c)a^3+(c-a)b^3$$

whereupon one will find that

$$P(a)=a+b+c$$

Thus,

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

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    @dxiv Thanks for the catch! – 2017-01-12