Let $a,b\in G$ elements of order $5$. If $a^3=b^3$ then $a=b$. 
Prove or disprove: let $G$ be a group and $a,b\in G$ elements of order $5$. If $a^3=b^3$ then $a=b$.

I saw the following example which tries to disprove the theorem: $G=\mathbb{Z}_{10}$ and $a=2,b=8$.
I'm not sure about that part, but $o(2)=5$ and $o(2)=8$. I think that $o(2)=\infty$ because there is not $n$ so $2^n\,mod\,10=1$ but I'm not sure.
Does this example disproves the theorem?

No, it is true, the special case $\,m,n = 3,5\,$ below.
Lemma $\ \gcd(m,n) = 1,\ a^m = b^m,\ a^n = b^n\,\Rightarrow\, a = b$
Proof $\ $ The set $\,S\,$ of $\,k\in \Bbb Z\,$ such that $\,a^k = b^k\,$ is closed under subtraction hence, by  a 1-line proof, its least positive element $\,d\,$ divides every element, so $\,d\mid m,n\,$ coprime, thus $\,d =1.$
