## Intermediate Algebra for College Students (7th Edition)

$x^4y^4\sqrt{y}$
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, $\sqrt{(x^4y^4)^2(y)}=x^4y^4\sqrt{y}$