#### Answer

$x^4y^4\sqrt{y}$

#### Work Step by Step

Factor the radicand (expression inside the radical sign) so that at least one factor is a square to obtain:
$=\sqrt{x^8y^8(y)}
\\=\sqrt{(x^4y^4)^2(y)}$
The principal square root is always non-negative.
However, the value of $x^4y^4$ will never be negative, so we do not need to add absolute value bars.
Thus,
$\sqrt{(x^4y^4)^2(y)}=x^4y^4\sqrt{y}$