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 $4$.

Let $t=\sin x$ and solve the cubic $$t^3+t^2+t=1$$
Wolfram Alpha gives the real solution as 
$$t=(1/3) (-1 - 2/(17 + 3 \sqrt {33})^{1/3} + (17 + 3 \sqrt{33})^{1/3})$$
Plug the real solution of the above to get  $$(1-t^2)^3 -4(1-t^2)^2+8(1-t^2) =4$$ 
