How do you evaluate #(-27/8)^(4/3)#? Algebra Exponents and Exponential Functions Fractional Exponents 1 Answer Alan P. · Hubert May 25, 2015 #(-27/8)^(4/3)# #=( (-27/8)^(1/3))^4# #= ((root(3)(-27))/root(3)(8))^4# #= (-3/2)^4# #= 81/16# Answer link Related questions What are Fractional Exponents? How do you convert radical expressions to fractional exponents? How do you simplify fractional exponents? How do you evaluate fractional exponents? Why are fractional exponents roots? How do you simplify #(x^{\frac{1}{2}} y^{-\frac{2}{3}})(x^2 y^{\frac{1}{3}})#? How do you simplify #((3x)/(y^(1/3)))^3# without any fractions in the answer? How do you simplify #\frac{a^{-2}b^{-3}}{c^{-1}}# without any negative or fractional exponents... How do you evaluate #(16^{\frac{1}{2}})^3#? What is #5^0#? See all questions in Fractional Exponents Impact of this question 4086 views around the world You can reuse this answer Creative Commons License
