How do you long divide $\dfrac{{{x}^{4}}-{{y}^{4}}}{x-y}$?

Hint: The long division of polynomials is very much similar to the division of numbers. We will start by dividing the first term of the dividend ${{x}^{4}}-{{y}^{4}}$ by the divisor $x-y$ to get the first term of the quotient as ${{x}^{3}}$. Then we multiply the first term of the quotient ${{x}^{3}}$ with the divisor $x-y$ to get ${{x}^{4}}-{{x}^{3}}y$ and subtract it from the dividend ${{x}^{4}}-{{y}^{4}}$ to obtain the remainder as ${{x}^{3}}y-{{y}^{4}}$ which will be treated as the new dividend. Then we follow the similar procedure until we obtain the remainder equal to zero.

Complete step-by-step answer:
The division to be carried out is given in the question as $\dfrac{{{x}^{4}}-{{y}^{4}}}{x-y}$. Let us call this as $p\left( x \right)$ and so we can write the below equation
$p\left( x \right)=\dfrac{{{x}^{4}}-{{y}^{4}}}{x-y}$
From above, we can see that the dividend is equal to ${{x}^{4}}-{{y}^{4}}$ and the divisor is equal to $x-y$. Just like the division of numbers, we carry out the long division as below
\[x-y\overset{{{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}+{{y}^{3}}}{\overline{\left){\begin{align}
  & {{x}^{4}}-{{y}^{4}} \\
 & \underline{{{x}^{4}}-{{x}^{3}}y} \\
 & \underline{\begin{align}
  & {{x}^{3}}y-{{y}^{4}} \\
 & {{x}^{3}}y-{{x}^{2}}{{y}^{2}} \\
\end{align}} \\
 & \underline{\begin{align}
  & {{x}^{2}}{{y}^{2}}-{{y}^{4}} \\
 & {{x}^{2}}{{y}^{2}}-x{{y}^{3}} \\
\end{align}} \\
 & \underline{\begin{align}
  & x{{y}^{3}}-{{y}^{4}} \\
 & x{{y}^{3}}-{{y}^{4}} \\
\end{align}} \\
 & \underline{0} \\
\end{align}}\right.}}\]
Hence, we long divided $\dfrac{{{x}^{4}}-{{y}^{4}}}{x-y}$ and obtained the quotient as \[{{x}^{3}}+{{x}^{2}}y+x{{y}^{2}}+{{y}^{3}}\] and the remainder equal to $0$.

Note: In the long division method, chances of committing the mistakes are huge. So always check your result using the simple relation between the divisor, the dividend, and the remainder which states that the dividend is equal to sum of the remainder and the product of the divisor and the quotient. Also, in the case of the above question, we can clearly see that we can apply the identity ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ on the dividend ${{x}^{4}}-{{y}^{4}}$ to obtain it as ${{x}^{4}}-{{y}^{4}}=\left( x-y \right)\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)$. So the divisor is a factor of the dividend and so it will get cancelled and we will obtain the quotient as $\left( x+y \right)\left( {{x}^{2}}+{{y}^{2}} \right)$. So this can also be the method of checking our result in this case.