## Calculus 8th Edition

$x=\pm 1$ or $x=\pm i$
We solve by factoring: $x^{4}=1$ $x^{4}-1=0$ We use the fact that $a^2-b^2=(a+b)(a-b)$: $(x^{2}-1)(x^{2}+1)=0$ $x^{2}-1=0$ or $x^{2}+1=0$ $x^2=1$ or $x^2=-1$