QPaperGen

We use the identity:
\[
\cos 3A = 4 \cos^3 A - 3 \cos A
\]

Also, recall:
\[
\cos 2x = 2 \cos^2 x - 1
\]

Now consider the left-hand side:
\begin{align*}
\cos 6x &= \cos 3(2x) \\
&= 4 \cos^3(2x) - 3 \cos(2x)
\end{align*}

Substitute \(\cos 2x = 2 \cos^2 x - 1\):
\begin{align*}
\cos 6x &= 4 \left( 2 \cos^2 x - 1 \right)^3 - 3 \left( 2 \cos^2 x - 1 \right)
\end{align*}

Now expand the cube:
\[
\left( 2 \cos^2 x - 1 \right)^3 = 8 \cos^6 x - 12 \cos^4 x + 6 \cos^2 x - 1
\]

So:
\begin{align*}
\cos 6x &= 4 \left( 8 \cos^6 x - 12 \cos^4 x + 6 \cos^2 x - 1 \right) - 3 \left( 2 \cos^2 x - 1 \right) \\
&= 32 \cos^6 x - 48 \cos^4 x + 24 \cos^2 x - 4 - 6 \cos^2 x + 3 \\
&= 32 \cos^6 x - 48 \cos^4 x + 18 \cos^2 x - 1 \qquad \textbf{(Hence, proved.)}
\end{align*}