If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$ 
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$

My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\big(1+\sin^2x\big)\\
&=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\
&=
\end{align}
I don't think its getting anywhere with my attempt, so how do I solve it ?
Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?
Note: The solution given in my reference is $4$.

Write $s := \sin x, c := \cos x$.
Hint Using $s^2 + c^2 = 1$, we can write our expression as
$$c^6 - 4 c^4 + 8 c^2 = (1 - s^2)^3 - 4 (1 - s^2)^2 + 8 (1 - s^2) = -s^6 - s^4 - 3 s^2 + 5 .$$
Now, perform polynomial long division by $s^3 + s^2 + s - 1$.

 Carrying out long division gives that our expression is $$(s^3 + s^2 + s - 1) p(s) + 4$$ for some cubic polynomial $p$. But we're given that $s^3 + s^2 + s = 1$, so the first term is zero, and thus $$\color{#df0000}{\boxed{c^6 - 4 c^4 + 8 c^2 = 4}} .$$

