How do you simplify #((2x^-6)/(7y^-3))^-3#?

2 Answers
Jan 9, 2016

Answer:

#(343x^18)/(8y^9)#

Explanation:

A negative power #x^(-n)# can be seen as #x^(n^(-1))#. We know that #x^(-1) = 1/x# so #x^(n^(-1)) = 1/x^n#, hence the following simplyfictations : #2x^-6 = 2/x^6# and #7y^-3 = 7/y^3#

So #((2x^-6)/(7y^-3))^(-3) = ((2/x^6)/(7/y^3))^(-3) = ((2/x^6)*(y^3/7))^(-3)#.

We do what we did before : #((2/x^6)*(y^3/7))^(-3) = (x^6/2*7/y^3)^3 = (343x^18)/(8y^9)#

Jan 9, 2016

Answer:

A slightly different way or writing the working out!

#(343x^18)/(8y^9) #

Explanation:

Given: #( (2x^(-6))/(7y^(-3)))^(-3)#

First consider inside the brackets:

Known that #color(white)(..)2x^(-6) -> 2/(x^6)" and that " 1/(7y^(-3)) ->y^3/7#

So we have #2/x^6xxy^3/7= (2y^3)/(7x^6)#

Now put this back into the brackets giving

#((2y^3)/(7x^6))^(-3) = ((7x^6)/(2y^3))^3#

#=(343x^18)/(8y^9) #