#### Answer

$x=\pm1$

#### Work Step by Step

$x^{8}+15x^{4}=16$
$x^{8}+15x^{4}-16=0$
$(x^{4}+16)(x^{4}-1)=0$
$x^{4}+16=0$ or $x^{4}-1=0$
$x^4=-16$ or $x^4=1$
$x=\pm\sqrt[4]{-16}$ or $x=\pm\sqrt[4]{1}=\pm1$
The left solution is not real, so we ignore it. Thus the solution is $x=\pm1$