### Instruction: If $a$, $b$ are nonzero real numbers such that $a^2+b^2=8ab$, find the value of $\left|\frac{a+b}{a-b}\right|$.### Response: We are given that $a^2+b^2=8ab$. So, $a^2-2ab+b^2=6ab$.
That means that $(a-b)^2=6ab$.
And $|a-b|=\sqrt{6ab}$.
So, $\left|\frac{a+b}{a-b}\right|=\frac{|a+b|}{|a-b|}=\frac{|a+b|}{\sqrt{6ab}}$.
We also have that $a^2+b^2=8ab$, so $(a+b)^2=10ab$.
That means that $|a+b|=\sqrt{10ab}$.
So, $\left|\frac{a+b}{a-b}\right|=\frac{|a+b|}{|a-b|}=\frac{\sqrt{10ab}}{\sqrt{6ab}}$.