Solve the equation $(1-x)\sqrt{1-x^4}=x$ 
Solve the equation
  $$(1-x)\sqrt{1-x^4}=x$$

My work so far:
$$(1-x)^2(1-x^4)=x^2$$
$$(1-2x+x^2)(1-x^4)=x^2$$
$$1-x^4-2x+2x^5+x^2-x^6=x^2$$

You are completely correct, we obtain the polynomial equation
$$
x^6 - 2x^5 + x^4 + 2x - 1=0.
$$
The polynomial is irreducible over $\mathbb{Q}$, i.e., it does not factor into factors of smaller degree. We have exactly two real roots of the degree $6$ polynomial equation, namely
$x=0.492425875905$ and $x=- 0.935635630515$. The second one is not a solution of the original equation.
