Simplify.

`root(4)(32x^11y^15)/root(4)(2x^3y^2)`

` `

### 2 Answers | Add Yours

We know

`x^m/y^m=(x/y)^m`

Thus apply this rule

`root(4)(32x^11y^15)/root(4)(2x^3y^2)=root(4)((32x^11y^15)/(2x^3y^2))`

`=(16x^(11-3)y^(15-2))^(1/4)`

`=(2^4x^8y^13)^(1/4)`

`=(2^4)^(1/4)(x^8)^(1/4)(y^13)^(1/4)`

`=2x^2y^3(y)^(1/4)`

``

`root(4)(32x^11y^15)/root(4)(2x^3y^2)=` `root(4)((32x^11y^15)/(2x^3y^2))` `=root(4)(16x^8y^13)` `=root(4)(2^4(x^2)^4 (y^3)^4 y)` `=2x^2y^3root(4)(y)`

### Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes