How do you simplify #(-i)^3#? Precalculus Complex Numbers in Trigonometric Form Powers of Complex Numbers 2 Answers Annie Nov 12, 2016 #i# Explanation: #(-i)^3# =#(-i)# x #(-i)# x #(-i)# But #(-i)# x #(-i)# =#+i ^2# = -1 So #(-i)^3# =#(-i)# x #(-i)# x #(-i)#= #-i # x -1= #i # Answer link seol · Stefan V. Nov 12, 2016 #i# Explanation: #i# is #sqrt(-1)#. Therefore, #i^2# would be #-1#... #-i^3=-i \times (-i) \times (-i)=-1 \times (-i) =i# 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 7595 views around the world You can reuse this answer Creative Commons License