#### Answer

$(x^{2}+y)^{4}$

#### Work Step by Step

We need to factor:
$x^{8}+4x^{6}y+6x^{4}y^{2}+4x^{2}y^{3}+y^{4}$
We rewrite this as:
$(x^2)^4+4(x^2)^3y^1+6(x^2)^2y^2+4(x^2)^1y^3+y^4$
We notice that this corresponds to an expanded binomial with a power of $4$ and terms $x^2$ and $y$:
$(x^{2}+y)^{4}$