Answer
1, -1 , $\frac{1+i\sqrt3}{2}$, $\frac{1-i\sqrt3}{2}$
Work Step by Step
$x^4-x^3+x-1=x^3(x-1)+1(x-1)=(x^3+1)(x-1)=0$
$(x+1)(x^2-x+1)(x-1)=0$
Either x+1=0 giving x=-1
or x-1=0 giving x=1
or $x^2-x+1=0$ giving x = $\frac{1+i\sqrt3}{2}$ or $\frac{1-i\sqrt3}{2}$
(using the discriminant formula for roots of the quadratic equation, x = $ \frac{-b+\sqrt{b^2-4ac}}{2a}$ or $ \frac{-b-\sqrt{b^2-4ac}}{2a}$)
