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I was just asked to factor $x^3+1$. I came to $(x^2-x+1)(x+1)$ after a while, and I was wondering, whether there is a good method to quicky factor such of polynomials.

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Look for a simple root $a$ of the polynomial and divide the initial polynomial by $x-a$. –  Beni Bogosel Nov 28 '12 at 20:30
    
There is no nice way to factor most cubics. But school assignment cubics usually have at least one rational root, which can be quickly located. –  André Nicolas Nov 28 '12 at 20:35
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For any odd $\,n\in\Bbb N\,$ , we get:

$$a^n+b^n=(a+b)(a^{n-1}-a^{n-2}b+a^{n-3}b^2-...-ab^{n-2}+b^{n-1})$$

Note that if $\,n\,$ is even the above does not work (the alternating signs in the right parentheses don't fix as one would expect!)

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Have a look at the rational root theorem and polynomial division.

This comes in very handy!

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You are usually asked to remember $x^3-a^3=(x-a)(x^2+ax+a^2)$ and $x^3+a^3=(x+a)(x^2-ax+a^2)$. In general, $$x^n-a^n=(x-a)(x^{n-1}+ax^{n-2}+...+a^{n-2}x+a^{n-1})$$

If $$y=x^{n-1}+ax^{n-2}+...+a^{n-2}x+a^{n-1}$$ $$xy=x^n+ax^{n-1}+...+a^{n-2}x^2+a^{n-1}x$$ $$-ay=-ax^{n-1}-a^2x^{n-2}-...-a^{n-1}x-a^n$$

Add $xy$ and $-ay$ and all the terms will cancel out except for $x^n-a^n$, showing

$$xy-ay=(x-a)y=x^n-a^n$$

For your equation, $n=3$ and $a=-1$.

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