Show that $\cos^220^\circ-\cos20^\circ\sin10^\circ+\sin^210^\circ=\frac34$ The original exercise is to

Prove that
$$4(\cos^320^\circ+\sin^310^\circ)=3(\cos20^\circ+\sin10^\circ)$$

Dividing both sides by $\cos20^\circ+\sin10^\circ$ leads me to the problem in the question title.
I've tried rewriting the left side in terms of $\sin10^\circ$:
$$4\sin^410^\circ+2\sin^310^\circ-3\sin^210^\circ-\sin10^\circ+1\quad(*)$$
but there doesn't seem to be any immediate way to simplify further. I've considered replacing $x=10^\circ$ to see if there was some observation I could make about the more general polynomial $4x^4-2x^3-3x^2-x+1$ but I don't see anything particularly useful about that. Attempting to rewrite in terms of $\cos20^\circ$ seems like it would complicate things by needlessly(?) introducing square roots.
Is there a clever application of identities to arrive at the value of $\dfrac34$? I have considered
$$\cos20^\circ\sin10^\circ=\frac{\sin30^\circ-\sin10^\circ}2=\frac14-\frac12\sin10^\circ$$
which eliminates the cubic term in $(*)$, and I would have to show that
$$4\sin^410^\circ-3\sin^210^\circ+\frac12\sin10^\circ=0$$
$$4\sin^310^\circ-3\sin10^\circ+\frac12=0$$

We need $$4\cos^220^\circ-4\cos20^\circ\sin10^\circ+4\sin^210^\circ=3$$
$$\iff2(1+\cos40^\circ)-2(\sin30^\circ-\sin10^\circ)+2(1-\cos20^\circ)=3$$
$$\iff\cos40^\circ-\cos20^\circ=-\sin10^\circ$$
which is evident from Prosthaphaeresis Formulas
