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

1 Answer
Aug 16, 2016

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

Explanation:

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#