Does $3^{1/2} \in \mathbb Q(3^{1/3}+3^{1/2})$?
I did try creating the minimal polynomial of $3^{1/3}+3^{1/2}$ which was of degree $6$ which turned out to be $f(x)=x^6 - 9 x^4 - 6 x^3 + 27 x^2 - 54 x - 18$. Though I'm not entirely sure if this was the aim of the question. Is there any other method?
Consider $\gamma=3^{1/3}+3^{1/2}$.
Then $(\gamma-3^{1/2})^3=3$. So $\gamma^3-3\gamma^23^{1/2}+9\gamma-3\cdot 3^{1/2}=3$.
Thus we can rearrange that into, $$\gamma^3+9\gamma-3=3\gamma^23^{1/2}+3\cdot 3^{1/2}$$
$$\dfrac{\gamma^3+9\gamma-3}{3\gamma^2+3}=3^{1/2}$$
So we see that $3^{1/2}$ is a rational function of $\gamma$. So $3^{1/2}\in\mathbb{Q}(\gamma)$. Which also shows that $3^{1/3}\in\mathbb{Q}(\gamma)$.