Question

How do you find the value of `( \frac { 216} { 1000} ) ^ { \frac { 2} { 3} }`?

Answer 1

`= 9/25`

Explanation:

One of the laws of indices states:

`rootq x^p = x^(p/q)`

So, `(216/1000)^(2/3)` means the same as `root3(216/1000)^2`

Note that `216 and 1000` are both perfect cubes.

`216 = 6^3 and 1000 = 10^3`

`root3((216/1000))^2" "larr` find the cube roots first

`= (6/10)^2" "larr` simplify the fraction

`=(3/5)^2" "larr` square the fraction

`= 9/25`

Answer 2

`9/25`

Explanation:

`( \frac { 216} { 1000} ) ^ { \frac { 2} { 3} }`

Solution

`( \frac { 216} { 1000} ) ^ { \frac { 2} { 3} }` `rArr` `3sqrt(216/1000)^2`

Cube root of `216 = 6`

Cube root of `1000 = 10`

`:.` `(6/10)^2 = 36/100 = cancel36^9/cancel100_25 = 9/25 -> Answer`

Answer 3

`color(magenta)(9/25`

Explanation:

`(216/1000)^(2/3)`

`:.=((3^3*cancel(2^3)^1)/(5^3*cancel(2^3)^1))^(2/3)`

`:.=3^(6/3)/(5^(6/3))`

`:.=3^2/5^2`

`:.=color(magenta)(9/25`

Another method:

`(a/b)^(2/3)=root3((a/b)^2)`

`:.=root3((216/1000)^2)`

`:.=root3(46656/1000000)`

`:.=root3((3*3*3*3*3*3*cancel2*cancel2*cancel2*cancel2*cancel2*cancel2)/(5*5*5*5*5*5*cancel2*cancel2*cancel2*cancel2*cancel2*cancel2))`

`:.root3(a)*root3(a)*root3(a)=a`

`:.=3^2/5^2`

`:.color(magenta)(=9/25)`