help

So if the equation was ((-125)^4)/3 then you would raise -125 to the fourth power which would be 244140625. Then you would divide by three and get 81380208.3333

If the expression is: (-125)^(4/3)

Simplify the following:
(-125)^(4/3)

(-125)^(4/3) = (-125)^(3/3+1/3) = (-125)^(3/3)×(-125)^(1/3):
(-125)^(3/3) (-125)^(1/3)

3/3 = 1:
-125 (-125)^(1/3)

(-125)^(1/3)  =  (-1×125)^(1/3)  =  (-1×5^3)^(1/3):
-125 (-5^3)^(1/3)

(-5^3)^(1/3)  =  (-1)^(1/3) (5^3)^(1/3)  =  (-1)^(1/3)×5^(3/3)  =  (-1)^(1/3)×5:
-125(-1)^(1/3)×5

-125×5  =  -625:
Answer: |   -625 (-1)^(1/3)

( - 125)^ 4 / 3 as a simplified fraction

I think our guest made a small error

\(( - 125)^ {4 / 3} \\ =( - 125)^ {1+1/3} \\ =( - 125)^ 1\times( - 125)^{1/3} \\ =\; - 125\times\;-5 \\ =625 \)

or alternatively

\(( - 125)^ {4 / 3} \\ =(( - 125)^ {1/3})^4 \\ =(-5)^4 \\ =625 \)

Melody: Please look at this Wolfram/Alpha result and see if you can explain it. Thanks:

http://www.wolframalpha.com/input/?i=simplify++%28-125%29^%284%2F3%29

Hi Guest

No I cannot explain how they got that answer.

Perhaps if I was a subscriber and had access to the Wolfram|alpha working I would understand it but alas I do not.

Fractional powers of negative numbers always can be interpreted in different ways giving different answers.

Also the cubic root of an number will always have 3 answers.  One of them may by real but the other 2 will be complex.

however.

If you press on   "Assuming the principal root | Use the real‐valued root instead"

which is in the initial lines of the Wolfram|Alpha answer, you will see my answer of 625 appear.   :)