Suppose $G$ is a group that for every $a,b\in G, (ab)^4=a^4b^4$. Prove that $N$ is normal subgroup if $N=G^3$ 
Suppose $G$ is a group that for every $a,b \in G$, $(ab)^4=a^4b^4$. Prove that $N$ is normal subgroup if $N=G^3=\{g^3\mid g\in G\}$

My attempt: We need to show that $gg^3g^{-1} \in N$
$gg^3g^{-1} = g(ggg)g^{-1}$ for the associative law we get $ gg(ggg^{-1})$
since $gg^{-1}=e$ we get $ggge=ggg=g^3 \in N$
But I could be completely wrong.
 A: Let $h\in G$ be arbitrary. The assumption $(gh)^4=g^4 h^4$ implies that $N=G^3$ is a group, since for arbitrary $g,h\in G$ one has $(gh)^4=g(hg)^3h=g^4h^4$, and so $g^3h^3=(hg)^3$. Also from $(gh)^4=g^4h^4$ you can conclude that
$$h^{-1}g^3h = ghghgh^{-2}=ghghgh (h^{-1})^3=(gh)^3(h^{-1})^3\in N,$$
and this shows that $N$ is normal.

Thanks to @DerekHolt and @ArturoMagidin for pointing out that the approach is overcomplicated. The first few lines that show that $N$ is a group are enough to conclude that it is a normal subgroup, since clearly $(h^{-1}gh)^n=h^{-1}g^nh$, or as Arturo points out, $G^k$ will be invariant under any endomorphism of $G$ for any $k$.
A: I will use the one-step subgroup test.
Since $e^3=e\in N$, we have $N\neq\varnothing 
$.
Let $n\in N$. Then there is some $m\in G$ such that $n=m^3$. But $m^3\in G$, so $n\in G$. Hence $N\subseteq G$.
Let $a,b\in N$. Then $a=g^3, b=h^3$ for some $g,h\in G$. Now
$$\begin{align}
ab^{-1}&=g^3(h^{-1})^3\\
&=g^3h^{-3}\\
&=(h^{-1}g)^3\\
&\in N
\end{align}$$
since $$(gh^{-1})^4=g^4h^{-4}\tag{1}$$ implies $(h^{-1}g)^3=g^3h^{-3}$ by multiplying $(1)$ on the left by  $g^{-1}$ and on the right by $h$.
Hence $N\le G$.
It remains to show that $N$ is normal in $G$. To do this, it suffices to show that for each $g\in G$ and each $n\in N$, we have $gng^{-1}\in N$. So fix $g\in G, n\in N$, so that there is some $h\in G$ with $n=h^3$; but then
$$\begin{align}
gng^{-1}&=gh^3g^{-1}\\
&=(ghg^{-1})(ghg^{-1})(ghg^{-1})\\
&=(ghg^{-1})^3,
\end{align}$$
but $ghg^{-1}\in G$ by closure. Hence $gng^{-1}\in N$.
Hence $N\unlhd G$.
