How do you simplify #root3(y^6)/(root3(27y)*root3(y^11))#?

1 Answer
Feb 22, 2017

Answer:

#1/(3y^2)#

Explanation:

Because all the terms are cube rooted (and the expression contains no addition or subtraction), the cube root can be moved to the outside of the expression: #root3(y^6/(27y.y^11))#

We can simplify the denominator using the fact that #a^n.a^m=a^(n+m)#
Both terms in the denominator are #y# raised to a power, so when we multiply them we add the indices to get:

#root3(y^6/(27y^12))#

We can now divide the top and the bottom of the fraction by #y^6# giving us #1# in the numerator, as #y^6/y^6=1#

For the denominator we use #a^n/a^m=a^(n-m)# to get: #27y^12/y^6=27y^(12-6)=27y^6#

So the expression becomes: #root3(1/(27y^6))#

The cube root of #27# is #3#.

#root3(x)# is equivalent to #x^(1/3)#, so #root3(y^6)=y^(6/3)=y^2#

Therefore the expression fully simplifies to: #1/(3y^2)#