$$16 { x }^{ 4 } -8 { x }^{ 2 } { y }^{ 2 } + { y }^{ 4 }$$
Factor
$\left(2x-y\right)^{2}\left(2x+y\right)^{2}$
Solution Steps
Consider $16x^{4}-8x^{2}y^{2}+y^{4}$ as a polynomial over variable $x$.
$$16x^{4}-8y^{2}x^{2}+y^{4}$$
Find one factor of the form $kx^{m}+n$, where $kx^{m}$ divides the monomial with the highest power $16x^{4}$ and $n$ divides the constant factor $y^{4}$. One such factor is $4x^{2}-y^{2}$. Factor the polynomial by dividing it by this factor.
Consider $4x^{2}-y^{2}$. Rewrite $4x^{2}-y^{2}$ as $\left(2x\right)^{2}-y^{2}$. The difference of squares can be factored using the rule: $a^{2}-b^{2}=\left(a-b\right)\left(a+b\right)$.
