How do you evaluate # 48^(4/3)*8^(2/3)*(1/6^2)^(3/2) #? Prealgebra Exponents, Radicals and Scientific Notation Exponents 1 Answer Shwetank Mauria May 2, 2016 #48^(4/3)* 8^(2/3)*(1/6^2)^(3/2)=2^(13/3)/3^(1/3)# Explanation: #48^(4/3)* 8^(2/3)*(1/6^2)^(3/2)# = #(2xx2xx2xx2xx3)^(4/3)* (2xx2xx2)^(2/3)*((2xx3)^(-2))^(3/2)# =#(2^4xx3)^(4/3)* (2^3)^(2/3)*(2xx3)^(-2xx3/2)# = #2^(4xx4/3)*3^(4/3)*2^(3*2/3)*(2xx3)^(-3)# = #2^(16/3)* 3^(4/3)* 2^2*2^(-3)*3^(-3)# = #2^(16/3+2-3)xx3^(4/3-3)# = #2^(13/3)xx3^(-1/3)# = #2^(13/3)/3^(1/3)# Answer link Related questions How do you simplify #c^3v^9c^-1c^0#? How do you simplify #(- 1/5)^-2 + (-2)^-2#? How do you simplify #(4^6)^2 #? How do you simplify #3x^(2/3) y^(3/4) (2x^(5/3) y^(1/2))^3 #? How do you simplify #4^3ยท4^5#? How do you simplify #(5^-2)^-3#? How do you simplify and write #(-5.3)^0# with positive exponents? How do you factor #12j^2k - 36j^6k^6 + 12j^2#? How do you simplify the expression #2^5/(2^3 times 2^8)#? When can I add exponents? See all questions in Exponents Impact of this question 2005 views around the world You can reuse this answer Creative Commons License