Question

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

Answer 1

`(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)`

Answer 2

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)`