True of false: For every integer $\displaystyle n \geq 2,$ there exists a non-abelian group of order $\displaystyle n^3.$
If $\displaystyle n$ is even, $\displaystyle D_{n^3/2}$ is nonabelian of order $\displaystyle n^3$.
If $\displaystyle n$ is odd, perhaps we can decompose $\displaystyle n$ into a product of prime powers and show that for every odd prime $\displaystyle p$ there exists a nonabelian group of order $\displaystyle p^3$...
It is sufficient to prove the result for $\displaystyle n=p$, a prime number, as then cross products give us the result.
So, a non-abelian group of order $\displaystyle p^3$? Does the group with presentation $\displaystyle G=<x,a|x^p=1, a^{p^2}=1, a^x=a^{1+p}>$ work? Clearly it is non-abelian (as if it was $\displaystyle a^x=a \neq a^{1+p}$), and it has order $\displaystyle p^3$ (as $\displaystyle |G| = o(g_1) * \ldots * o(g_i)$ with $\displaystyle G=<g_1, \ldots, g_m>$ a minimal generating set and $\displaystyle o(g_i) < \infty$).
(This is a specific case of the group $\displaystyle G=<x,a|x^p=1, a^{p^{n-1}}=1, a^x=a^{1+p^{n-2}}>$, $\displaystyle n \geq 3$, the only non-abelian p-group that has a cyclic maximal subgroup and is not of maximal class - see Robinson, A Course in the Theory of Groups, section 5.3.4)