How do you find the cube roots of #1000#? Trigonometry The Polar System De Moivre’s and the nth Root Theorems 1 Answer Shwetank Mauria Mar 1, 2017 Cube roots of #1000# are #10#, #(-5+5sqrt3i)# and #(-5-5sqrt3i)# Explanation: Cube roots of #1# are well known namely #1#, #-1/2+sqrt3/2i# and #-1/2-sqrt3/2i# Hence cube roots of #1000=10^3# are #10xx1=10#, #10xx(-1/2+sqrt3/2i)=-5+5sqrt3i# and #10xx(-1/2-sqrt3/2i)=-5-5sqrt3i# Answer link Related questions How do you use De Moivre’s Theorem to find the powers of complex numbers in polar form? What is the DeMoivre's theorem used for? How do you find the #n^{th}# roots of complex numbers in polar form? How do you find #[2(\cos 120^\circ + i \sin 120^\circ)]^5# using the De Moivre's theorem? What is #(-\frac{1}{2}+\frac{i\sqrt{3}}{2})^{10}#? How do you find the three cube roots of #-2-2i \sqrt{3}#? If the roots can be determined, will some form of De Moivre’s Theorem be used? How do you find the two square roots of 2i? Question #f6317 How do you find #z, z^2, z^3, z^4# given #z=sqrt2/2(1+i)#? See all questions in De Moivre’s and the nth Root Theorems Impact of this question 7701 views around the world You can reuse this answer Creative Commons License