How do you find all solutions to #x^3+1=0#?

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Answer 1 · 2016-08-16T20:05:56

#x = -1 or 1/2 +- (sqrt(3))/2i#

Using synthetic division and the fact that #x=-1# is obviously a solution we find that we can expand this to:

#(x+1)(x^2-x+1) = 0#

In order to have LHS=RHS need one of the brackets to be equal to zero, ie

#(x+1) = 0" " color(blue)(1)#

#(x^2-x+1) = 0" " color(blue)(2)#

From #1# we note that #x = -1# is a solution. We shall solve #2# using the quadratic formula:

#x^2-x+1 = 0#

#x = (1+-sqrt((-1)^2-4(1)(1)))/2 = (1+-sqrt(-3))/2 = (1+-sqrt(3)i)/2#