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 $x_2=-1/2-isqrt3/2$

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