How do you simplify #root3 (x^6 y^7)/root3 9#?

1 Answer
May 10, 2018

#root(3)(x^6y^7)/root(3)9# can be written as =#(xy)^2root(3)(y/9)#

Explanation:

You have the expression #root(3)(x^6y^7)/root(3)9#
Remember that the cubic root #root(3)# of something means a number which multiplied with itself three times #(z*z*z)# gives the number under the root sign.

("multiplied with itself three times" is, of course, not quite precise, but I mean that n is a factor three times, i.e. #z*z*z#.)

Let's rewrite the expression a little:

#root(3)(x^6y^7)/root(3)9 = root(3)((xy)^6y/3^2)#
=#(xy)^2root(3)(y/3^2)#
Or if you will
=#(xy)^2root(3)(y/9)#
depending on what you prefer.

We may be fooled to think that #root(3)9# can be simplified, but it is not a cubic number. If we had had #27=3^3# instead, we would get the nice expression
=#(xy)^2/3root(3)(y)#