How do you simplify #root3(125x^21y^24)#?

1 Answer
May 11, 2016

Answer:

#root(3)(125x^21y^24) = 5x^7y^8#

Explanation:

Notice that #(5x^7y^8)^3 = 5^3(x^7)^3(y^8)^3 = 125x^21y^24#

So #5x^7y^8# is a cube root of #125x^21y^24#

Any Real number has a unique Real cube root, so for Real values of #x# and #y# and Real cube roots, we can deduce that this is the cube root we want.