Question

How do you write `root3(x^-6)` with fractional exponent?

Answer 1

`x^-2`

Explanation:

Note that

`roota(x^b)=x^(a//b)`

Thus,

`root3(x^-6)=x^(-6//3)=x^-2`

Note that

`x^-2=1/x^2`

both of which are acceptable answers.

Answer 2

Very slight expansion on the process

`1/x^2 = x^(-2)`

Explanation:

Given:`" "root(3)(x^(-6))`

Consider the `x^(-6)` this is the same as `1/x^6`

Put it back into the root giving

`" "root(3)(1/x^6)`

This is the same as

`" " (root(3)(1))/(root(3)(x^6))`

But `x^6 = x^(2+2+2) = x^2xx x^2xx x^2 = (x^2)^3` giving

`" "(root(3)(1))/(root(3)((x^2)^3)) = 1/x^2`

`"'~~~~~~~~~~~~ Important Note ~~~~~~~~~~~~~~~~~~~~~~~"`

The key question is about the state being positive or negative.

Consider `root(3)(1)`: for 1 to be negative there would need to be an indicator in the starting point that complex numbers are involved. This is not the case so this question has no credence.

Consider the `x^2xx x^2xx x^2`: if `x` where to be negative then each `x^2` would change it back to positive. So it is possible for `x<1`, that is, the source value could be either negative or positive.

However the given expression has been simplified such that it involves `x^2`. This automatically results in an outcome that is positive.

So it is correct to write the answer as positive. That is, we do not have negative `1/x^2`