Using FLT for $ n=3$ Using FLT for exponent $3$, I need to show that if n positive integer is divisible by $3$, then there are no $x,y,z$ positive integers such that $x^n+y^n=z^n$.
This is what I did: 
 if $3/n$ then $n=3.k$, for some $k$ positive integer. Then if $(x^3, y^3,z^3)$ is a solution for the exponent $k$ 
$(x^3)^k+(y^3)^k=(z^3)^k$
and because $3.k=k.3$ in integers 
$(x^k)^3+(y^k)^3= (z^k)^3$
where $x^k=a ,y^k,=b z^k=c$ are positive integers. Then there are no , $a,b,c$ positive integers such that $a^3+b^3=c^3$ .
Could this be remotely correct?

if $x^{3k} + y^{3k} = z^{3k}$ then set $u=x^k$, $v=y^k$, $w=z^k$ and we have $u^{3} + v^{3} = w^{3}.$
Therefore if $u^{3} + v^{3} = w^{3}$ has no nontrivial solutions, $x^{3k} + y^{3k} = z^{3k}$ has no nontrivial solutions.
