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I am stuck on a lecture problem.

Prove that if $x\in\mathbb{Z}$ and $x<0$, then $x^3<0$.

I believe that I am supposed to use the integer axioms but am having trouble making the first step or which ones to use. Any help would be appreciated. Thank You!

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    Note, a negative number times a negative number is positive and a positive number times a negative number is negative. Finally note that $x^3=(x\cdot x)\cdot x$. If you haven't been given "negative times negative is positive" to work with yet, then prove that first. – 2017-02-21
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    Can you provide a link to the axioms with which you can work; alternatively, it would help if you could edit your post to include those axioms. – 2017-02-21

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HINT If $x < 0$ then let $x = -y$ with $y > 0$ and now $$ x^3 = (-y)^3 = (-y) \times (-y) \times (-y) = y^2 \times(-y) = -y^3... $$

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    .. or use $x^3+y^3=(x+y)(x^2+xy+y^2)=0$ – 2017-02-21