Given that $f(x^3)=[f(x)]^3$ for $f:K\to K'$ with $f(x+y)=f(x)+f(y)$, $f(1)=1$ and $K$ and $K'$ two fields with characteristic not equal$2$ and $3$. Given that $f(x^3)=[f(x)]^3$ for $f:K\to K'$ with $f(x+y)=f(x)+f(y)$, $f(1)=1$  and $K$ and $K'$ two fields with characteristic different from $2$ and $3$ show that $f$ is ring homomorphism. 
My Attempt: For a given $x\in K$ observe that $$f((1+x)^3)=[f(1+x)]^3\implies 3f(x^2)=3[f(x)]^2.$$  I somehow want to conclude from this that $f(x^2)=[f(x)]^2$ but I am not sure how to do so. Maybe this involves using the fact that $\text{char}(K')\not =3.$ It would be great if someone could provide an argument that completes the proof of this observation.

Having $f((1+x)^3)=(1+f(x))^3\implies 1+3f(x^2)+3f(x)+f(x^3)=1+3f(x)^2+3f(x)+f(x)^3$, we have $f(x^2)=f(x)^2$, since $char~K\neq 3$. Then applying $f((x+y)^2)=(f(x)+f(y))^2$, we have $f(xy)=f(x)f(y)$, since $char~K\neq 2$
