Question 203101
{{{root(3, (108x^3*y^7)/(2y^3))}}}
Let's start by reducing the fraction inside:
{{{root(3, (108x^3*y^7)/(2y^3)) = root(3, ((2*54*x^3*y^3*y^4)/(2*y^3))) = root(3, ( cross(2)*54*x^3* cross(y^3)*y^4)/( cross(2) * cross(y^3))) = root(3, 54x^3*y^4)}}}
Now we look for perfect cube factors:
{{{root(3, 54x^3*y^4) = root(3, 27*2*x^3*y^3*y) = root(3, 3^3*2*x^3*y^3*y)}}}
Now we can use one of the properties of roots (which apply for all roots: square, cube, 4th, 5th, etc.): {{{sqrt(x*y) = sqrt(x)*sqrt(y)}}}. This allows us to separate factors in a root into separate roots:
{{{root(3, 3^3*2*x^3*y^3*y) = root(3, 3^3)*root(3, 2)*root(3, x^3)*root(3, y^3)*root(3, y)}}}
Now we can simplify the cube roots of the perfect cubes:
{{{root(3, 3^3)*root(3, 2)*root(3, x^3)*root(3, y^3)*root(3, y) = 3*root(3, 2)*x*y*root(3, y)}}}
Using the Commutative Property of multiplication to rearrange the order:
{{{3* root(3, 2)*x*y* root(3, y) = 3xy*root(3, 2)*root(3, y)}}}
And finally, using the same property of roots as earlier (except in reverse), to combine the cube roots:
{{{3xy*root(3, 2)*root(3, y) = 3xy*root(3, 2y)}}}