If sin⁡ x+sin² x+sin³ x=1 then cos⁶x-4cos⁴ x+8cos² x=

The given equation is $\sin x+{\sin }^{2}x+{\sin }^{3}x=1$

This can be rewrite as $\sin x\left(1+{\sin }^{2}x\right)={\cos }^{2}x$

it implies that $\sin x\left(2-{\cos }^{2}x\right)={\cos }^{2}x$

Squaring on both sides

 ${\sin }^{2}x\left(4+{\cos }^{4}x-4{\cos }^{2}x\right)={\cos }^{4}x$

it implies that $\left(1-{\cos }^{2}x\right)\left(4+{\cos }^{4}x-4{\cos }^{2}x\right)={\cos }^{4}x$

$\begin{array}{rcl}4+{\cos }^{4}x-4{\cos }^{2}x-4{\cos }^{2}x-{\cos }^{6}x+4{\cos }^{4}x& =& {\cos }^{4}x\\ {\cos }^{6}x-4{\cos }^{4}x+8{\cos }^{2}x& =& 4\end{array}$

Therefore, the value of the given expression is $\boxed{\mathbf{4}}$