Question 1036509

square root of 128x^3 over the square root of 2xy. thank you!!! im not really sure what im doing, i was absent in class. 
<pre>{{{sqrt(128x^3)/sqrt(2xy)}}} =====> {{{sqrt(128x^3/(2xy))}}} =====> {{{sqrt(64cross(128)x^2cross(x^3)/(cross(2)cross(x)y))}}} =====> {{{sqrt(64x^2/y)}}} =====> {{{highlight_green(matrix(1,3, 8x/sqrt(y), or, 8x*sqrt(y)/y))}}}