Show that $\cos(6x)= 32\cos^6x -48\cos^4x +18\cos^2x -1$ After writing down $\cos6x$= $Re (\cos x + i\sin x)^6$, I used the binomial theorem to expand the expression. Very soon it got really tedious and after trying $5$ times, fruitlessly, to arrive at the given expression, I gave up. Is there a shorter way around this?

Perhaps a little bit less tedious then a binomial expansion to the power of $6$, you could use the triple and double angle identities $\cos 3x=4\cos^3x-3\cos x$ and $\cos 2x=2\cos^2x-1$, so that we have
$$\begin{align}\cos (6x)&=\cos(3\cdot 2x)\\&=4\cos^3(2x)-3\cos(2x)\\&=4(2\cos^2x-1)^3-3(2\cos^2x-1)\\&=4(8\cos^6x-12\cos^4x+6\cos^2x-1)-6\cos^2x+3\\&=32\cos^6x-48\cos^4x+18\cos^2x-1\end{align}$$
