Trigonometric analytical proof How would you go about showing that

$-4\cos^{3}(t) +3\cos(t) = -\cos(3t)$

It's part of a larger problem, and I know that to be true, but I just can't figure out how to get there (I might be too tired, though).
Thanks for your help!
Write $\mathrm{cos}(3x)$ as $\mathrm{cos}(2x+x)$, use the formula for the cosine of a sum, and insert $\mathrm{cos}(2x)=\mathrm{cos}^2(x)-\mathrm{sin}^2(x)$ and $\mathrm{sin}(2x)=2\mathrm{sin}(x)\mathrm{cos}(x)$.
$$ Cos(3t) = Cos(2t+t) \\ = cos(2t)cos(t) - sin(t)sin(2t) \\ = [{cos^2(t)} - {sin^2(t)}] cos(t) - sin(t) [2sin(t) cos(t)] \\ = [{cos^2(t)} - [ 1- {cos^2(t)}]cos(t) - 2{sin^2(t) cos(t)} \\ = [cos^2(t) - 1 + cos^2(t)]cos(t) - 2cos(t) [1-cos^2(t)] \\ = 2cos^3(t) - cos(t) - 2cost + 2cos^3(t) \\ = 4cos^3(t) - 3cos(t) \\ \text {Now  we  know} \\
Cos(3t) = 4cos^3(t) - 3cos(t) \\ \text {so multiply both sides by -1} \\
-Cos(3t) = 3cos(t) - 4cos^3(t)
$$