How do you solve #a^3 = 4a #? Algebra Quadratic Equations and Functions Comparing Methods for Solving Quadratics 1 Answer Shwetank Mauria Sep 27, 2016 #a=0# or #a=-2# or #a=2# Explanation: #a^3=4a# #hArra^3-4a=0# or #a(a^2-4)=0# or #a(a^2-2^2)=0# or #a(a+2)(a-2)=0# Hence either #a=0# or #a+2=0# or #a-2=0# i.e. #a=0# or #a=-2# or #a=2# Answer link Related questions What are the different methods for solving quadratic equations? What would be the best method to solve #-3x^2+12x+1=0#? How do you solve #-4x^2+4x=9#? What are the two numbers if the product of two consecutive integers is 72? Which method do you use to solve the quadratic equation #81x^2+1=0#? How do you solve #-4x^2+4000x=0#? How do you solve for x in #x^2-6x+4=0#? How do you solve #x^2-6x-16=0# by factoring? How do you solve by factoring and using the principle of zero products #x^2 + 7x + 6 = 0#? How do you solve #x^2=2x#? See all questions in Comparing Methods for Solving Quadratics Impact of this question 1339 views around the world You can reuse this answer Creative Commons License