Solving trigonometric identity with condition. Problem : If $\sin\theta +\sin^2\theta +\sin^3\theta=1$ Then prove $\cos^6\theta -4\cos^4\theta +8\cos^2\theta =4$
My working :
As $\sin\theta +\sin^2\theta +\sin^3\theta=1 \Rightarrow \sin\theta +\sin^3\theta = \cos^2\theta$
Now the given equation : $\cos^6\theta -4\cos^4\theta +8\cos^2\theta$ can be written as 
$(\sin\theta +\sin^3\theta)^3-4(\sin\theta+\sin^3\theta )^2+8(\sin\theta +\sin^3\theta)$
= $\sin^3\theta +\sin^6\theta +3\sin^5\theta +3\sin^7\theta -4\sin^2\theta -4\sin^6\theta -8\sin^4\theta + 8\sin\theta + 8\sin^3\theta$
But I think this is not the right way of doing this...Please suggest other alternative.. Thanks...

Let $x = \sin \theta$. We are given that $x^3 + x^2 + x - 1 = 0$.
We want to show that $(1-x^2)^3 -4(1-x^2)^2 +8(1-x^2) = 4$. Expanding and comparing terms, this is equivalent to 
$$ x^6 + x^4 + 3x^2 - 1 = 0. $$
This is true because $$x^6 + x^4 + 3x^2 - 1 = (x^3 + x^2 + x -1 ) ( x^3 - x^2 + x +1).$$
