How do you simplify #(\frac { 27x ^ { 9} } { 8} ) ^ { \frac { 5} { 3} }#?

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Answer 1 · 2018-02-25T12:21:48

# (243x^15)/32#

# ((27x^9)/8)^(5/3) = (((27x^9)/8)^(1/3))^5 #

#((3x^3)/2)^5 = (243*x^15)/32#

#= (243x^15)/32# [Ans]

Answer 2 · 2018-02-25T12:40:16

# (243x^15)/32#

In #((27x^9)/8)^(5/3)# it will help if you notice that all the factors are cubes.

We can write this as #((3^3x^9)/(2^3))^(5/3)#

The fraction in the exponent can be written as a cube root and an index:

#((3^3x^9)/(2^3))^(5/3) = root3(((3^3x^9)/(2^3)))^5#

#= ((3x^3)/2)^5#

#= (243x^15)/32#