Algebra

16x^8y^2)^1/2

sqrt(16x^8y^2)

4x^?y

 ~  what is the square root of x^8?

since  

$x^4*x^4=x^8\\ \sqrt{x^8}=x^4$

$\sqrt{16x^8y^2}\\ =\sqrt{4*4*x^4*x^4*y*y}\\ =\sqrt{(4*x^4*y)\;\;\;*\;\;\;(4*x^4*y)}\\ =4*x^4*y\\ =4x^4y$