How do you simplify #sqrt(-75)^3#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 1 Answer Somebody N. Apr 16, 2018 #color(blue)(i375sqrt(3))# Explanation: #sqrt((-75)^3)=sqrt(-1)*sqrt(75^3)# #:.# #sqrt(75^3)=(75)^(3/2)=(sqrt(75))^3=(sqrt(3xx25))^3=(5sqrt(3))^3# #=5^3(sqrt(3))^3=3*5^3sqrt(3)=375sqrt(3)# #:.# #375sqrt(3)*sqrt(-1)=i375sqrt(3)# #color(blue)(i375sqrt(3))# Answer link Related questions How do I use DeMoivre's theorem to find #(1+i)^5#? How do I use DeMoivre's theorem to find #(1-i)^10#? How do I use DeMoivre's theorem to find #(2+2i)^6#? What is #i^2#? What is #i^3#? What is #i^4#? How do I find the value of a given power of #i#? How do I find the #n#th power of a complex number? How do I find the negative power of a complex number? Write the complex number #i^17# in standard form? See all questions in Powers of Complex Numbers Impact of this question 3344 views around the world You can reuse this answer Creative Commons License