Answer
$\log_{a}[x^{6}-x^{4}y^{2}+x^{2}y^{4}-y^{6}]$
Work Step by Step
$x^{8}-y^{8}$ is a difference of squares,
$x^{8}-y^{8}=(x^{4}-y^{4})(x^{4}+y^{4})$
$x^{4}-y^{4}$is a difference of squares,
$x^{8}-y^{8}=(x^{2}+y^{2})(x^{2}-y^{2})(x^{4}+y^{4})$
Applying the rule: $\log_{a}(MN)=\log_{a}M+\log_{a}N$
$\log_{a}(x^{8}-y^{8})=\log_{a}[(x^{2}-y^{2})(x^{2}+y^{2})]+\log_{a}(x^{2}+y^{2})$
So,
$\log_{a}(x^{8}-y^{8})-\log_{a}(x^{2}+y^{2})=\log_{a}[(x^{2}-y^{2})(x^{4}+y^{4})]$
$=\log_{a}[x^{6}-x^{4}y^{2}+x^{2}y^{4}-y^{6}]$
