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when I factor $x^3-1$ with $\mod 3$ on maple I get the answer $(x+2)^3$ and I was just wondering what the steps were to get to this solution.

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    You can [post mathematical expressions](http://math.stackexchange.com/help/notation) here using MathJax and $\LaTeX$. One thing to understand is that $2\equiv -1 \bmod{3}$. – 2017-02-01
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    It’s quite easy, when you remember that in characteristic $p$ (prime), you have the identity $(X+Y)^p=X^p+Y^p$. Here, the $X$ is $x$, and the $Y$ is $-1\equiv2\pmod3$. – 2017-02-01

4 Answers 4

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\begin{align*} x^3-1 &\equiv x^3-3x^2+3x-1\,(\text{mod}\,\, 3)\\ &\equiv (x-1)^3\\ &\equiv (x+2)^3\,(\text{mod}\,\, 3) \end{align*}

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We have that $$x^3-1=(x-1)(x^2+x+1)\equiv (x-1)(x^2-2x+1)=(x-1)^3\equiv(x+2)^3 \pmod{3}.$$

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\begin{align*} x^3 - 1 &= (x - 1)(x^2 + x + 1)\\[6pt] &\equiv (x + 2)(x^2 + 4x + 4) \pmod{3}\\[6pt] &\equiv (x + 2)(x + 2)^2 \pmod{3}\\[6pt] &\equiv (x + 2)^3 \pmod{3} \end{align*}

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$$x^3-1$$ $$= (x-1)(x^2+x+1)$$ $$= (x-1)(x^2+x - 3x +1)$$ $$= (x-1)(x-1)^2$$ $$= (x-1)^3$$ $$= (x+2)^3$$

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    so the root of this would be 1 with a multiplicity of 3? – 2017-02-02
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    Root of this in Z/3Z – 2017-02-02