Question

Simplify `(125/729)^(-1/3)`?

Answer 1

`(125/729)^(-1/3)=9/5`

Explanation:

We should use the identities `(a^m)^n=a^(mxxn)`, `a^(-m)=1/a^m` and `a^(1/m)=root(m)a`

Hence, `(125/729)^(-1/3)`

= `((5xx5xx5)/(3xx3xx3xx3xx3xx3))^(-1/3)`

= `(5^3/3^6)^(-1/3)`

= `5^((3xx(-1/3)))/3^((6xx(-1/3))`

= `5^(-1)/3^(-2)`

= `(1/5)/(1/9)`

= `1/5xx9/1`

= `9/5`

Answer 2

`9/5`

Explanation:

Recall: `x^-m = 1/x^m" and " (color(red)(x)/color(blue)(y))^-m = (color(blue)(y)/color(red)(x))^m`

I usually try to get rid of any negative indices first.
Use the second law shown above to do this.

`(125/729)^(-1/3) " = "(729/125)^(1/3)`

It is useful to learn all the powers up to 1000.

You should recognise these values as being cubes.
(`5^3 = 125 and 9^3 = 729`)

Recall: `root3(x) hArr x^(1/3)`

`(729/125)^(1/3) = root3((729/125)) = (root3 729)/(root3 125) = (root3 (9^3))/(root3 (5^3))`

`=9/5`