<s>[INST] 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
$$4\sin^410^\circ-3\sin^210^\circ+\frac12\sin10^\circ=0$$
$$4\sin^310^\circ-3\sin10^\circ+\frac12=0$$
 [/INST] A complex numbers proof of the original exercise.
By Euler's formula, we have that
$$\text{Re}\left((\cos 20^\circ+i\sin 20^\circ)^3\right)=\cos(60^\circ)=\frac{1}{2}=\sin(30^\circ)=\text{Im}\left((\cos 10^\circ+i\sin 10^\circ)^3\right)$$
Hence
$$\cos^3 20^\circ-3\cos 20^\circ \sin^2 20^\circ
=3\cos^2 10^\circ\sin 10^\circ -\sin^3 10^\circ$$
or
$$\cos^3 20^\circ-3\cos 20^\circ (1-\cos^2 20^\circ)
=3(1-\sin^2 10^\circ)\sin 10^\circ -\sin^3 10^\circ
$$
which implies
$$4(\cos^320^\circ+\sin^310^\circ)=3(\cos20^\circ+\sin10^\circ).$$
The very same argument leads to the following generalization: if $c^\circ +s^\circ=30^\circ$ then
$$4(\cos^3 c^\circ+\sin^3 s^\circ)=3(\cos c^\circ+\sin s^\circ).$$
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