For every $g\in G$ there exists an $h\in G$ such that $g = h^3$ Let G be a finite group whose order is not divisible by $3$. Show that for every $g\in G$ there exists an $h\in G$ such that $g = h^3$.
 A: Let $a$ be the order of $g$. Then $a$ divides the order of $G$, and therefore $a$ is not divisible by $3$.  Because $a$ and $3$ are relatively prime, by Bezout's Theorem there exist integers $x$ and $y$ such that $ax+3y=1$. Without loss of generality we can assume that $y \ge 0$. It follows that
$$g=g^1=g^{ax+3y}.$$
Can you finish from here?
Remark: One does not really need Bezout's Theorem in this case, since it is easy to prove the existence of suitable $x$ and $y$ with essentially no machinery. But if we replace $3$ by a large prime $p$, the machinery would be needed.  
Explicitly, suppose that $a$ has remainder $2$ on division by $3$. Let $h=g^{(a+1)/3}$. Then $h^3=g$.
Suppose that $a$ has remainder $1$ on division by $3$. Let $h=g^{(2a+1)/3}$. Then $h^3=g$.  
Proving these two facts is just a computation. For the first one, $h^3=g^{a+1}=g^ag=eg=g$. for the second one, the computation is very similar. 
We used the more complicated approach of the post so that you could see how to generalize.
A: Hint $\rm\ If\,\ g^n\! = 1,\ gcd(3,n)=1\:$ then, by Bezout, $\rm\:mod\ n\!:\ J\equiv\dfrac{1}3\:$ exists, so $\rm\ g\, =\, g^{\,3\,J} =\, (g^J)^3\ \ $
Remark $\ $ The key idea is that, because $\rm\: g^n = 1,\:$ exponents on $\rm\,g\,$ can be calculated modulo $\rm n,\:$ where integers $\rm\:k\:$ coprime to $\rm\:n\:$ are invertible. Therefore $\rm\:g\:$ has a $\rm\,k$'th root: $\rm\: g^{\,1/k\ mod\ n},\ $ or, written additively, $\rm\: \left(\frac{1}k\ mod\ n\right)\cdot g.$
