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

Given that $\sin x+{\sin }^{2}x+{\sin }^{3}x=1$

$\sin x+{\sin }^{3}x={\cos }^{2}x$

Squaring on both sides

${\sin }^{2}x+{\sin }^{6}x+2{\sin }^{4}x={\cos }^{4}x$

$(1-{\cos }^{2}x)+{(1-{\cos }^{2}x)}^3+2{(1-{\cos }^{2}x)}^2={\cos }^{4}x$

$1-{\cos }^{2}x+1-{\cos }^{6}x-3{\cos }^{2}x+3{\cos }^{4}x+2(1+{\cos }^{4}x-2{\cos }^{2}x)={\cos }^{4}x$

$4-{\cos }^{6}x+4{\cos }^{4}x-8{\cos }^{2}x=0$

${\cos }^{6}x-4{\cos }^{4}x+8{\cos }^{2}x=4$