If $k=\Bbb Q(\omega)$ and $K=k(\sqrt[3]2)$ then the isomorphism $K\otimes_k K\to K^3$ goes as follows: $$(a+b\sqrt[3]2+c\sqrt[3]4)\otimes\xi\mapsto ((a+b\sqrt[3]2+c\sqrt[3]4)\xi, (a+b\omega\sqrt[3]2+c\omega^2\sqrt[3]4)\xi, (a+b\omega^2\sqrt[3]2+c\omega\sqrt[3]4)\xi)$$ where $a$, $b$, $c\in k$, $\xi\in K$.
I want to show that this is isomorphism for this I need kernel to be trivial say tensor product map to (0,0,0) then we need to show that tensor product is 0 and then we need to show surjectivity how to show that? Also I want to know how K looks like is it same as Q($\omega$,$\sqrt[3]2$)?
I'm not sure what $\omega$ is but from context, I will assume its a third root of unity, so that $\mathbb{Q}(\omega, 2^{\frac{1}{3}})= K$ is the splitting field of $x^3-2$. Then we have that $$ K \otimes_k K = K \otimes_k k[x]/(x^3-2) = K[x]/(x^3-2) = K^3 $$ since $x^3-2$ splits completely over $K$.