Chapter 1, Equations and Graphs - Section 1.6 - Solving Other Types of Equations - 1.6 Exercises - Page 138: 23

$x=2$

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 $x=2$