$3$ does not divide $|G| \implies \forall \ g \in G, g=h^3$ , for some $h$ in $G$ 
If $G$ be a finite group such that $3$ does not divide the order of  $G$.
  Then show that every element of $G$ can be expressed as $h^3$ , for some $h \in G$.

Thought:  I have just figured out that if $g$ has order $3$ then $g= h^3$ is not possible. But nothing else.
 A: Hint: if a number $a$ is not divisible by $3$ then there exists a number $b$ such that $3 b \equiv 1 \mod a.$
A: Let $n$ be the order of $g$, so $g^n = 1$, then $n | |G|$ by Lagrange, so $\gcd(3,n) = 1$ and we have $a,b \in \mathbb{Z}$ with $an + 3b = 1$. Then 
$$g = g^1 = g^{3b +an} = (g^b)^3 (g^n)^a  =(g^b)^3 1^a = (g^b)^3$$
so that $h = g^b$ is as required. 
A: Below is an alternate proof not using congruence. It only uses Lagrange's theorem and the definition of the order of an element. ($o(g)$ is defined as the least positive integer $k$ such that $g^k=1$.)

Since $|G|$ is not divisible by $3$, Lagrange's theorem says that $G$ has no elements of order $3$ (which implies $|G|$ not divisible by $3$ by Cauchy's theorem).
For every $g\in G$, the order of the subgroup $\left<g\right>=\{g^n\mid n=0,1,\cdots\}$ generated by $g$ is equal to the order $o(g)$ of $g$. Now we claim that $o(g^3)=o(g)$. Denote $m=o(g)$.
Clearly $(g^3)^m=(g^m)^3=e$ so $o(g^3)\le m$.
Conversely, since $(g^{o(g^3)})^3=(g^3)^{o(g^3)}=e$ and $G$ has no elements of order $3$, the order of $g^{o(g^3)}$ is $<3$. But if $(g^{o(g^3)})^2=e$, then
$g^{o(g^3)}=(g^{o(g^3)})^3\cdot((g^{o(g^3)})^2)^{-1}=e$. So the order of $g^{o(g^3)}$ is $1$. Namely, $g^{o(g^3)}=e$. This means $m\le o(g^3)$. Thus $o(g^3)=o(g)$.
Now the two subgroups $\left< g\right>$ and $\left<g^3\right>$ are of the same order and evidently $\left<g^3\right>\subseteq\left< g\right>$.
Therefore $\left<g^3\right>=\left< g\right>$. This means $g=g^{3n}$ for some $n$. Namely, $g=(g^n)^3$. Q.E.D.

Hope this helps.
