How do you solve #x^3-1=0#?

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Answer 1 · 2016-11-30T19:43:17

The solutions are #S={1, -1/2+isqrt3/2,-1/2-isqrt3/2} #

We use #(a^3-b^3)=(a-b)(a^2+ab+b^2)#

Therefore.

#(x^3-1)=(x-1)(x^2+x+1)=0#

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

#x=1#

For the equation, #x^2+x+1=0#

We calculate the discriminant,

#Delta=b^2-4ac=1-4*1*1=-3#

As, #Delta>0#, we do not a solution in #RR# but in #CC#

#x=(-1+-sqrtDelta)/2=(-1+-isqrt3)/2#

#x_1=-1/2+isqrt3/2# and #x_2=-1/2-isqrt3/2#