Question

Is it okay to have fractional exponents such as `r^(2/3)` in your answer?

I am supposed to simplify the cube root of `-64r^2x^15`. I have gotten as far as `-4x^5 root(3)(r^2)`. Is this as far as I need to go?

Answer 1

It is better to not use `r^(2/3)` as `-4x^5root(3)(r^2)` is already the simplest form and may actually mean something different from `-4x^5r^(2/3)`.

Explanation:

Given:

`-64r^2x^15`

we can note that:

`(-4x^5)^3 = (-4)^3 (x^5)^3 = -64x^(5 * 3) = -64x^15`

So `-4x^5` is a cube root of `-64x^15`.

Real cube roots of real numbers are unique, since `f(t) = t^3` is a one to one function of real numbers.

So `root(3)(-64x^15) = -4x^5`

Similarly, if `root(3)(r^2)` is the cube root of `r^2` then

`(-4x^5root(3)(r^2))^3 = -64x^15 (root(3)(r^2))^3 = -64x^15r^2`

and we can deduce that:

`root(3)(-64x^15r^2) = -4x^5root(3)(r^2)`

One interesting thing to note here is that for any real value of `r`, we have `r^2 >= 0` and hence `root(3)(r^2)` is understood to be a non-negative real number.

What about `r^(2/3)` ?

Whatever it is, it is certainly a cube root of `r^2` in that

`(r^(2/3))^3 = r^((2/3)*3) = r^2`

What about if `r = -1` ?

It is clear enough what we intend by:

`root(3)((-1)^2) = root(3)(1) = 1`

but what does `(-1)^(2/3)` actually mean?

In the context of complex numbers, it can be taken to stand for the principal value:

`(-1)^(2/3) = cos (pi/3) + i sin (pi/3) = 1/2+sqrt(3)/2i`

Some people prefer to use fractional expressions as multi-valued when dealing with complex or negative numbers.

So in at least some cases, the expression `r^(2/3)` is not the same as `root(3)(r^2)`.