Finding $a,b,c$ using the equation provided Provided with the equation
$$\sin^2 (n+1)x - \sin^2 (n-1)x = \sin2nx \sin2x$$
Find the values of $a,b,c$ if
$$ \sin^2 3x - \sin^2 x = 8\cos^2 x (a\cos^4 x + b\cos^2 x + c)$$
What I have tried
$(1 - \cos^2 3x)- (1- \cos^2 x)=8\cos^2 x (a\cos^4 x + b\cos^2 x + c)$
Expanding out the LHS
$(1 - (\cos 2x \cos x - \sin 2x \sin x )^2)  + \cos^2 x -1 $
$=( \cos x( 2\cos^2 x -1)-2\sin^2 x \cos x)^2$
$=\cos^2 x((2\cos^2x -1) -2(1-\cos^2 x))^2 $
$=\cos^2 x(4\cos^2 x -3)^2 $
$=16\cos^6 x -24\cos^4 x  + 9\cos^2 x$
By comparing,
$a=2$
$b=-3$
$c=\frac{9}{8}$
Is there a way to do the question using the equation they provided? I can't seem to see it.Any tips would be useful.Thank you!

Friend, your answer is wrong, 
Correct answer is  $a=-2$, $b=3$ and $c=-1$
For a quick verification, put $x=0$ ;
LHS = $0$
thus RHS must be $0$, i.e. $a+b+c=0$. Which isn't true in your case.
Proper proof using given equation :
$\sin^2(n+1)x−\sin^2(n−1)x=\sin2nx\sin2x$
Put $n=2$
$\sin^23x−\sin^2x= \sin4x\sin2x$
$= 2\sin^22x\cos2x$
$= 2(1-(2\cos^2x-1)^2)(2\cos^2x-1)$
$= 2(4\cos^2x-\cos^4x)(2\cos^2x-1)$
$= 8\cos^2x(1-\cos^2x)(2cos^2x-1)$
$= 8\cos^2x(-2\cos^4x+3\cos^2x-1)$
Thus ;
$a=-2$ ,
$b=3$ , and 
$c=-1$
