Question 354980
{{{x^3=1}}} Start with the given equation.



{{{x^3-1=0}}} Subtract 1 from both sides.



{{{(x-1)(x^2+x+1)=0}}} Factor the left side using the <a href="http://www.purplemath.com/modules/specfact2.htm">difference of cubes formula</a>.



{{{x-1=0}}} or {{{x^2+x+1=0}}} Use the zero product property.



Now solve {{{x-1=0}}} to get {{{x=1}}}



Next, solve {{{x^2+x+1=0}}} to get {{{x = (-1+i*sqrt(3))/(2)}}} or {{{x = (-1-i*sqrt(3))/(2)}}}. See <a href="http://www.algebra.com/algebra/homework/quadratic/Quadratic_Equations.faq.question.354990.html">this page</a> for the steps on solving {{{x^2+x+1=0}}}



So the solutions are {{{x=1}}}, {{{x = (-1+i*sqrt(3))/(2)}}} or {{{x = (-1-i*sqrt(3))/(2)}}} 



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Jim