To prove that $\frac{1}{n^3}$ is Cauchy you apply $\big|a_n-a_m\big|$ so it looks like $\big|\frac{1}{n^3} - \frac{1}{m^3}\big|$ but then you have to do something before applying the triangle rule and I'm not sure what.
Apparently $\big|\frac{1}{n^3}\big| + \big|\frac{1}{m^3}\big|<\epsilon \Rightarrow \frac{\epsilon}{2}=\frac{1}{N^3}\Rightarrow N=\big\lceil\sqrt[3]{\frac{2}{\epsilon}}\big\rceil$ is wrong.
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On the real line $\mathbb{R}$ you have the following
Theorem. A sequence converges if and only if it is Cauchy.
So if you can show that the sequence $\frac{1}{n^{3}}$ converges, then you are done.
You have some options now. Either show directly by the definition of convergence, that the sequence converges to a point, in this case $0$. Or notice that $\frac{1}{n} = n^{-3} = n^{-1} n^{-1} n^{-1}$ and show first that $n^{-1} = \frac{1}{n}$ converges to a point, in this case $0$.
Your original approach will work provided $n\in\mathbb N\setminus\{0\}$.
Given any $\epsilon>0$, you would like to find an $N$ such that for any $n,m>N$, $\left|\frac{1}{n^3}-\frac{1}{m^3}\right|<\epsilon$. This would certainly be the case if $\frac{1}{n^3},\frac{1}{m^3}<\frac{\epsilon}{2}$ since $\left|\frac{1}{n^3}-\frac{1}{m^3}\right|\leq\left|\frac{1}{n^3}\right|+\left|\frac{1}{m^3}\right|$. Therefore you need to force $n,m>\sqrt[3]{\frac{2}{\epsilon}}$. Conveniently enough, you are allowed to choose whichever $N$ you like, so choosing $N>\sqrt[3]{\frac{2}{\epsilon}}$ will do the trick. The above constitutes the work you do beforehand, now the proof.
Claim: The sequence $\{\frac{1}{n^3}\}$ is Cauchy.
Proof: Let $\epsilon>0$ be given and let $N>\sqrt[3]{\frac{2}{\epsilon}}$. Then for any $n,m>N$, one has $0<\frac{1}{n^3},\frac{1}{m^3}<\frac{\epsilon}{2}$. Therefore, $\epsilon>\frac{1}{n^3}+\frac{1}{m^3}=\left|\frac{1}{n^3}\right|+\left|\frac{1}{m^3}\right|\geq\left|\frac{1}{n^3}-\frac{1}{m^3}\right|$. Thus, the sequence is Cauchy as was to be shown.