Question 960287
your expression is:


(-27)^(5/3)


this is the same as (the cube root of -27) raised to the third power, or the cube root of (-27 raised to the third power).


in algebra.com, cube root of x would be shown as root(3,x).


(-27)^(5/3) would be shown as root(3,x)^3 or as root(3,x^3)


both forms would get you the same answer.


you cannot take an even root of a negative number.


root(2,-27) is not allowed.
root(4,-27) is not allowed.


you can take an odd root of a negative number.


root(3,-27) is allowed.
root(5,j-27) is allowed.


the restriction on the even root because the solution is not real.


for example, what is the square root of -4?


the square root of 4 is either plus 2 or minus 2 because 2*2 = 4 and -2 * -2 is 4.


there is no real number that you can multiply by itself to get -4.


with the odd root, however, there is.


cube root of -27 is equal to -3 because -3 * -3 * -3 is equal to -27.


since you are dealing with the cube root, you can take the cube root of -27 and then raise it to the fifth power, or you can raise -3 to the fifth power and then take the cube root of it.  


(-27)^(5/3) is equal to (-3)^5 which is equal to -243.