Please help me calculate the following:

16^(-1/2)

(27)^(1/3) - with questions that have two brackets, only the first number has brackets on paper.

(27)^(-1/3)

(-27)^(1/3)

(-27)^(-1/3)

(-2)^(-3)

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- Jul 9th 2008, 06:46 PMprincess_anna57Powers - please help!
Please help me calculate the following:

16^(-1/2)

(27)^(1/3) - with questions that have two brackets, only the first number has brackets on paper.

(27)^(-1/3)

(-27)^(1/3)

(-27)^(-1/3)

(-2)^(-3) - Jul 9th 2008, 06:59 PMReckoner
- Jul 9th 2008, 07:16 PMprincess_anna57Thanks!
I got the first three, but I'm not sure about the last ones. What difference does it make if x is a negative? does it mean that it's undefined/does not exist?

- Jul 9th 2008, 07:23 PMReckoner
Well, if you try to take the square root (or any

*even*root) of a negative number, you will not get a real answer (because the square of any real number is nonnegative). However, you*can*take cube roots of a negative number, and in fact you can take the $\displaystyle n^{\rm th}$ root of a negative number as long as $\displaystyle n$ is odd.

Thus, $\displaystyle \sqrt[3]{-27} = -3$, for example, because $\displaystyle (-3)^3 = (-3)(-3)(-3) = -27$.