Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$ 
Show that $\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = \cos^32\alpha$

I have tried $\sin^3\alpha(3\sin\alpha - 4 \sin^3\alpha) = 3\sin^4\alpha - 4\sin^6\alpha$ and $\cos^3\alpha(4\cos^3\alpha - 3\cos\alpha) = 4\cos^6\alpha - 3\cos^4\alpha$ to give
$$\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha = 3\sin^4\alpha - 4\sin^6\alpha + 4\cos^6\alpha - 3\cos^4\alpha$$ 
I can't work out how to simplify this to $\cos^32\alpha$.
I also noticed that the LHS of the question resembles $\cos(A-B)$, but I can't figure a way of making that useful.
 A: \begin{align}
L.H.S=& \sin 3\alpha \sin \alpha\sin^2\alpha+\cos 3\alpha\cos \alpha \cos^2 \alpha\\
\ =& \frac{1}{2}\left(\sin 3\alpha \sin \alpha (1-\cos 2\alpha)+\cos 3\alpha\cos \alpha(1+\cos 2\alpha)\right)\\
\ =& \frac{1}{2}\left(\sin 3\alpha \sin \alpha+ \cos 3\alpha\cos \alpha\right)+\frac{1}{2}\left(\cos 3\alpha\cos \alpha-\sin 3\alpha \sin \alpha\right)\cos 2\alpha\\
\ =& \frac{1}{2}\cos(3\alpha-\alpha)+\frac{1}{2}\cos(3\alpha+\alpha)\cos 2\alpha\\
\ =& \frac{1}{2}\cos 2\alpha(1+\cos 4\alpha)\\
\ =& \frac{1}{2}\cos 2\alpha \cdot 2\cos^2 2\alpha\\
\ =& \cos^3 2\alpha\hspace{6cm}\Box
\end{align}
A: Let $\cos^2\alpha=a,\sin^2\alpha=b\implies a+b=1, a-b=\cos2\alpha$
$ \sin3\alpha\sin^3\alpha + \cos3\alpha\cos^3\alpha = $
$ = 3\sin^4\alpha - 4\sin^6\alpha + 4\cos^6\alpha - 3\cos^4\alpha = $
$ = 3b^2-4b^3+4a^3-3a^2=4(a^3-b^3)-3(a^2-b^2)=(a-b)\{4(a^2+ab+b^2)-3(a+b)\} $
Now, $4(a^2+ab+b^2)-3(a+b)=4\{(a+b)^2-ab\}-3=4(1-ab)-3=1-4ab$
$=1-4\cos^2\alpha\sin^2\alpha=1-(2\sin\alpha\cos\alpha)^2=1-\sin^22\alpha=\cos^22\alpha$
A: Using $$\cos3A=4\cos^3A-3\cos A,\sin3A=3\sin A-4\sin^3A,$$
$$4(\sin3\alpha \sin^3\alpha + \cos3\alpha \cos^3\alpha)$$
$$=\sin3\alpha (3\sin\alpha-\sin3\alpha) + \cos3\alpha (\cos3\alpha+3\cos\alpha)$$
$$= \cos^23\alpha-\sin^23\alpha +3(\cos3\alpha\cos\alpha+\sin3\alpha \sin\alpha)$$
$$=\cos6\alpha+3\cos(3\alpha-\alpha)$$
(using $\cos2A=\cos^2A-\sin^2A,\cos(A-B)=\cos A\cos B+\sin A\sin B$)
$$=\cos(3\cdot2\alpha)+3\cos2\alpha$$
$$ = 4\cos^32\alpha-3\cos2\alpha+3\cos2\alpha (\text{ again applying }\cos3A)$$
$$ = 4\cos^32\alpha $$
