Answer
$\left( {{x}^{2}}+4{{y}^{2}} \right)\left( x+2y \right)\left( x-2y \right)$.
Work Step by Step
Consider the expression.
${{x}^{4}}-16{{y}^{4}}$
The given expression can be written as:
${{x}^{4}}-16{{y}^{4}}={{\left( {{x}^{2}} \right)}^{2}}-{{\left( 4{{y}^{2}} \right)}^{2}}$
Apply the formula of difference of squares,
$\begin{align}
& {{x}^{4}}-16{{y}^{4}}=\left( {{x}^{2}}+4{{y}^{2}} \right)\left( {{x}^{2}}-4{{y}^{2}} \right) \\
& =\left( {{x}^{2}}+4{{y}^{2}} \right)\left( {{x}^{2}}-{{\left( 2y \right)}^{2}} \right)
\end{align}$
Again, apply the formula of difference of squares in the term $\left( {{x}^{2}}-{{\left( 2y \right)}^{2}} \right)$,
${{x}^{4}}-16{{y}^{4}}=\left( {{x}^{2}}+4{{y}^{2}} \right)\left( x+2y \right)\left( x-2y \right)$
Check,
Apply the FOIL method.
$\begin{align}
& \left( {{x}^{2}}+4{{y}^{2}} \right)\left( x+2y \right)\left( x-2y \right)=\left( {{x}^{2}}+4{{y}^{2}} \right)\left( x\cdot x-x\cdot 2y+2y\cdot x-2y\cdot 2y \right) \\
& =\left( {{x}^{2}}+4{{y}^{2}} \right)\left( {{x}^{2}}-2xy+2yx-4{{y}^{2}} \right) \\
& =\left( {{x}^{2}}+4{{y}^{2}} \right)\left( {{x}^{2}}-4{{y}^{2}} \right)
\end{align}$
Again, apply the FOIL method.
$\begin{align}
& \left( {{x}^{2}}+4{{y}^{2}} \right)\left( x+2y \right)\left( x-2y \right)=\left( {{x}^{2}}+4{{y}^{2}} \right)\left( {{x}^{2}}-4{{y}^{2}} \right) \\
& =\left( {{x}^{2}}\cdot {{x}^{2}}-{{x}^{2}}\cdot 4{{y}^{2}}+4{{y}^{2}}\cdot {{x}^{2}}-4{{y}^{2}}\cdot 4{{y}^{2}} \right) \\
& =\left( {{x}^{4}}-4{{x}^{2}}{{y}^{2}}+4{{x}^{2}}{{y}^{2}}-16{{y}^{4}} \right) \\
& ={{x}^{4}}-16{{y}^{4}}
\end{align}$
