From the wording, it appears that you are asked to solve the problem without using DeMoivre's Theorem!
Then you need information about $\cos 3x$ and $\sin 3x$ from another source. Luckily, you probably have all the tools needed.
For $\cos 3x$, write it as $\cos(2x+x)$, and use the usual cosine of a sum rule. We get $\cos 2x\cos x-\sin 2x\sin x$. Using the double-angle identities, you then get $(\cos^2 x-\sin^2 x)\cos x-2\sin^2 x\cos x$. I used the rule $\cos 2x=\cos^2 x-\sin^2 x$ because it fits in well with what you are asked to prove.
We conclude that $\cos 3x=\cos^3 x-3\sin^2 x\cos x$.
You will need similar information about $\sin 3x$. Be guided by the ultimate result you are looking for. But start with $\sin 3x=\sin(2x+x)=\sin 2x \cos x+\cos 2x\sin x$. Of course you will then use $\sin 2x=2\sin x\cos x$, and an appropriate identity for $\cos 2x$.