I'm currently working on the below problem:
Let $\alpha$ and $\beta$ be elements of an extension field $L$ of $\mathbb{Q}$.
$(a)$ Prove that if $\alpha$ is algebraic of degree $2$ over $\mathbb{Q}$ and $\beta$ is algebraic of degree $3$ over $\mathbb{Q}$, then $\alpha \cdot \beta$ is algebraic of degree at most $6$.
$(b)$ Show by example that $\alpha \cdot \beta$ can have degree $6$.
$(c)$ Show by example that $\alpha \cdot \beta$ can have degree less than $6$.
Here is my work so far:
$(a)$ $\alpha$ is algebraic of degree $2$ over $\mathbb{Q}$ $\Rightarrow$ $p(\alpha) = 0$, where $p$ is the degree $2$ minimal polynomial of $\alpha$ over $\mathbb{Q}$. $\beta$ is algebraic of degree $3$ over $\mathbb{Q}$ $\Rightarrow$ $q(\beta) = 0$, where $q$ is the degree $3$ minimal polynomial of $\beta$ over $\mathbb{Q}$.
How can I use this data to conclude that the minimal polynomial of $\alpha \cdot \beta$ over $\mathbb{Q}$ has degree at most $3 \cdot 2 = 6$ over $\mathbb{Q}$? I'm only used to working with the Tower Theorem in these types of situations, but I'm not sure how to use this here since we're dealing with $[\mathbb{Q}(\alpha \cdot \beta): \mathbb{Q}]$ rather than $[\mathbb{Q}(\alpha, \beta): \mathbb{Q}]$.
$(b)$ I think $\alpha = 2^{\frac{1}{3}}$ and $\beta = \sqrt{2}$ will do the trick here. Then $\alpha \cdot \beta = 2^{5/6}$ $\Rightarrow$ $x = 2^{5/6}$ gives $x^6 = 32$, so $x^6-32$ is the minimal polynomial (of degree $6$) of $\alpha \cdot \beta$ over $\mathbb{Q}$.
$(c)$ All examples I can think of play out as (b) did. I'm not sure how to construct an example so that the desired degree is less than $6$. I've tried cube roots and square roots, imaginary numbers, and cube roots of unity, but I'm still not able to find one.
Thanks!