The remainder when  $x^4-y^4$ is divided by  $x-y$ is

Let  $f\left(x\right)=x^4-y^4$ and  $g\left(x\right)=x-y$

Now,  $g\left(x\right)=0\Rightarrow x-y=0\Rightarrow x=y$

By remainder theorem, we know that when  $f\left(x\right)$ is divided by  $g\left(x\right)$, then the remainder is  $f\left(y\right)$. Now  $f\left(y\right)=y^4-y^4=0$