How do you simplify #(-27)^(-2/3)#?

Question

2599 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-03-12T20:00:35

#1/9#

Since #a^(m/n)=root(n)(a^m)#

and

#a^-m=1/(a^m)#

then

#(-27)^(-2/3)=(-1/27)^(2/3)=root(3)((-1/27)^2)#

Since #27=3^3#, you get

#root(3)((-1/3)^6)=(-1/3)^2=1/9#

Answer 2 · 2017-03-12T21:39:01

#1/9#

Consider the example #x^(-2/3)# this is the same as #1/x^(2/3)#

Using this as our guide:

Write #(-27)^(-2/3)# as #1/(-27)^(2/3)#

This is the same as #1/root(3)((-27)^2)#

But I have just spotted something!
Lets take advantage of the fact that #(-3)xx(-3)xx(-3)=-27#

So changing the order we do things, write as:

#1/(root(3)(-27))^2 = 1/((-3)^2) = 1/9#