#### Answer

$x=2$

#### Work Step by Step

We factor by grouping:
$x^{3}-x^{2}+x-1=x^{2}+1$
$x^{3}-x^{2}+x-1-x^{2}-1=0$
$x^{3}-2x^{2}+x-2=0$
$x^{2}(x-2)+(x-2)=0$
$(x-2)(x^{2}+1)=0$
$x-2=0$ or $x^{2}+1=0$
$x=2$ or $x=\pm\sqrt{-1}$
We ignore non-real solutions. So the only real solution is $x=2$