Answer
$\displaystyle \frac{x^{4}+y^{4}}{x+y}=x^{3} -x^{2}y+xy^{2}-y^{3}+\frac{2y^{4}}{x+y}$
Work Step by Step
Dividend: $x^{4}+y^{4}$
... order by the powers of xy so the sum of exponents is 4 (there are missing powers)
$x^{4}+y^{4} =x^{4}+0\cdot x^{3}y+0\cdot x^{2}y^{2}+0\cdot xy^{3}+y^{4}$
Perform long division, aiming at the leading exponent of x.
$ \begin{array}{llllllllll}
& & x^{3} & -x^{2}y & +xy^{2} & -y^{3} & & \color{red}{ \small{Quotient}} & & \\
& & -- & -- & -- & -- & -- & -- & & \\
x+y & ) & x^{4} & +0x^{3}y & +0\cdot x^{2}y^{2} & +0\cdot xy^{3} & +y^{4} & & & \\
& & x^{4} & +x^{3}y & & & & \color{red}{\leftarrow \small{ x^{2}(x+y) } } & & \\
& & -- & -- & & & & \color{red}{\leftarrow \small{ subtract } } & & \\
& & & -x^{3}y & +0\cdot x^{2}y^{2} & +0\cdot xy^{3} & +y^{4} & & & \\
& & & -x^{3}y & -x^{2}y^{2} & & & \color{red}{\leftarrow \small{-x^{2}y(x+y) } } & & \\
& & & -- & -- & & & \color{red}{\leftarrow \small{ subtract }} & & \\
& & & & x^{2}y^{2} & +0\cdot xy^{3} & +y^{4} & & & \\
& & & & x^{2}y^{2} & +xy^{3} & & \color{red}{\leftarrow \small{ xy^{2}(x+y) } } & & \\
& & & & -- & -- & & \color{red}{\leftarrow \small{ subtract } } & & \\
& & & & & -xy^{3} & +y^{4} & & & \\
& & & & & -xy^{3} & -y^{4} & \color{red}{\leftarrow \small{ -y^{3}(x+y) } } & & \\
& & & & & -- & -- & \color{red}{\leftarrow \small{ subtract } } & & \\
& & & & & & 2y^{4} & \color{red}{\leftarrow \small{ Remainder } } & &
\end{array} $
$\displaystyle \frac{x^{4}+y^{4}}{x+y}=x^{3} -x^{2}y+xy^{2}-y^{3}+\frac{2y^{4}}{x+y}$
