Prove $\cos6x=32\cos^{6}x-48\cos^{4}x+18\cos^{2}x-1 $ So far I've done this:
LHS $ =\cos^{2}3x-\sin^{2}3x$
$={(4\cos^{3}x-3\cos{x})}^2 -{(3\sin{x}-4\sin^{3}x)}^2$
$=16\cos^{6}x+9\cos^{2}x-24\cos^{4}x-9\sin^{2}x-16\sin^{6}x+24\sin^{4}x$
I can tell I'm going in the right direction but how should I proceed further?

EDIT I used the identity $\cos{2x}=2\cos^{2}x-1$ to solve it in a simpler way. viz.
LHS  $= 2\cos^{2}3x-1$
$=2{(4\cos^{3}x-3\cos{x})}^2-1$
$2(16\cos^{6}x+9\cos^{2}x-24\cos^{4}x)-1$
$=32\cos^{6}x+18\cos^{2}x-48\cos^{4}x-1$
Still thank you for the answers!
 A: Another way:
$$\cos(3\cdot2x)=4\cos^3(2x)-3\cos2x$$
Now use $\cos2x=2\cos^2x-1$
A: Simpler way!
Use the fact $\displaystyle (\cos(x)+i\sin(x))^6 = \cos(6x)+i\sin(6x)$.
Expand, use $\sin^2(x)=1-\cos^2(x)$ to simplify higher powers of sine in the real terms,  and then organizing by real terms, we get 
$\cos(6x)+i\sin(6x)=$
$=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)+32i\sin(x)\cos(x)^5-32i\sin(x)\cos(x)^3+6i\sin(x)\cos(x)-1$
Consider the real part.
This gives you $\cos(6x)=32\cos^6(x)-48\cos^4(x)+18\cos^2(x)-1$.
If you consider the imaginary part, you can isolate $i\sin(6x)$.
This also gives you $\sin(6x)=2\sin(x)\cos(x)^5-32\sin(x)\cos(x)^3+6\sin(x)\cos(x)$
A: Note that you can use the identity $\cos^2 x+\sin^2 x=1$(or alternatively, $\sin^2 x=1-\cos^2 x$). 
Take each side of the equation to the power of $2,3$ to get $$9 \sin^2 x=9-9\cos^2 x$$
$$24\sin^4 x=24-48 \cos x^2+24 \cos^4 x$$
$$16\sin^6 x=16-48 \cos^2 x+48 \cos^4 x-16\cos^6 x$$
Tedious, but it probably will work. 
A: Uain Prosthaphaeresis Formulas, $$\cos6x+\cos 2x=2\cos4x\cos2x\iff\cos6x=\cos2x(2\cos4x-1)$$
$\cos4x=2\cos^22x-1$ and $\cos2x=2\cos^2x-1$
A: Expand $(\cos(x)+i\sin(x))^6=\cos(6x)+i\sin(6x) $ and take raeal part of both sides
