Answer
$\displaystyle \frac{x^{5}+y^{5}}{x+y}= x^4-x^3y+x^2y^2-xy^3+y^4$
Work Step by Step
Dividend: $x^{5}+y^{5}$
... order by the powers of xy so the sum of exponents is $5$ (there are missing powers)
$x^{5}+y^{5} =x^{5}+0\cdot x^{4}y+0\cdot x^{3}y^{2}+0\cdot x^{2}y^{3}+0\cdot xy^{4} +y^{5}$
Perform long division, aiming at the leading exponent of x.
$ \begin{array}{llllllllll}
& & x^{4} & -x^{3}y & +x^{2}y^{2} & -xy^{3} & +y^{4} & & \color{red}{ \small{Quotient}} & \\
& & -- & -- & -- & -- & -- & -- & & \\
x+y & ) & x^{5} & +0\cdot x^{4}y & +0\cdot x^{3}y^{2} & +0\cdot x^{2}y^{3} & +0\cdot xy^{4} & +y^{5} & & \\
& & x^{5} & +x^{4}y & & & & & \color{red}{\leftarrow \small{ x^{4}(x+y) } } & \\
& & -- & -- & & & & & \color{red}{\leftarrow \small{ subtract } } & \\
& & & -x^{4}y & +0\cdot x^{3}y^{2} & +0\cdot x^{2}y^{3} & +0\cdot xy^{4} & +y^{5} & & \\
& & & -x^{4}y & -x^{3}y^{2} & & & & \color{red}{\leftarrow \small{-x^{3}y(x+y) } } & \\
& & & -- & -- & & & & \color{red}{\leftarrow \small{ subtract }} & \\
& & & & x^{3}y^{2} & +0\cdot x^{2}y^{3} & +0\cdot xy^{4} & +y^{5} & & \\
& & & & x^{3}y^{2} & +x^{2}y^{3} & & & \color{red}{\leftarrow \small{ x^{2}y^{2}(x+y) } } & \\
& & & & -- & -- & & & \color{red}{\leftarrow \small{ subtract } } & \\
& & & & & -x^{2}y^{3} & +0\cdot xy^{4} & +y^{5} & & \\
& & & & & -x^{2}y^{3} & -xy^{4} & & \color{red}{\leftarrow \small{ -xy^{3}(x+y) } } & \\
& & & & & -- & -- & & \color{red}{\leftarrow \small{ subtract } } & \\
& & & & & & xy^{4} & +y^{5} & & \\
& & & & & & xy^{4} & +y5 & \color{red}{\leftarrow \small{ +y^{4}(x+y) } } & \\
& & & & & & -- & -- & \color{red}{\leftarrow \small{ subtract } } & \\
& & & & & & 0 & & \color{red}{\leftarrow \small{ Remainder } } &
\end{array} $
$\displaystyle \frac{x^{5}+y^{5}}{x+y}= x^4-x^3y+x^2y^2-xy^3+y^4$
