How would i prove that cuberoot 2 is not rational?

Let cuberoot 2 be written in the form m/n, where m and n are coprime.

So then $\displaystyle m^3=2n^3$

Hence $\displaystyle m^3$ and thus m is divisible by 2

So there exists integer k such that m=2k

thus $\displaystyle 8k^3$ = $\displaystyle 2n^3$

Thus $\displaystyle n^3$ = $\displaystyle 4k^3$

Where do i go next? im stuck. Does it matter that its 4k^3 instead of say 2k^3? Can i just say that n^3 is thus even?