High powers or low powers?

There are two methods which I know. I'll be typing the easier one and which involves less amount of latex work(I'm very bad at latex)

This method is kind of shortcut. As every person knows what to do in most questions of competitive exams..."Take values, substitute!"

Let's consider the equation $x^2-3x+2=0$ . Here, $p=-3,q=2$ and $\alpha=1,\beta=2$ to make things easier.

Substituting in each of the options as well as in the equation required, $(p^2-q)(p^2-3q)$ satisfies.

Well, I'm giving the other method too...

$\alpha^4+\alpha^2\beta^2+\beta^4$ can be written as $(\alpha^2+\beta^2)^2-(\alpha\beta)^2$ for our convenience.

Now, $\alpha+\beta=-p$ and $\alpha\beta=q$ Therefore, $(\alpha\beta)^2=q^2$

$\implies$ $(\alpha+\beta)^2=p^2$ which means $(\alpha^2+\beta^2)^2=(-2q+p^2)$ = $(p^2-2q)^2$

We require the value of $(\alpha^2+\beta^2)^2-(\alpha\beta)^2$

Plugging the values, we get $(p^2-2q)^2-q^2$ $(p^2-2q+q)(p^2-2q-q)$

Which gives $(p^2-q)(p^2-3q)$ and which is our answer.