How do you find three cube roots of $-1$ ?

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Answer 1 · 2017-02-22T07:00:19

Three cube roots of $1$ are ${-1,1/2-sqrt3/2i,1/2+sqrt3/2i}$

Let $x$ be the cube root of $-1$ , then we have $x*3=-1$

or $x^3+1=0$

$hArrx^3+x^2-x^2-x+x+1=0$

or $x^2(x+1)-x(x+1)+1(x+1)=0$

or $(x+1)(x^2-x+1)=0$

Hence either $x+1=0$ i.e. $x=-1$ , or $x^2-x+1=0$ .

So one root is $x=-1$ and for other two roots of $x^2-x+1=0$ , we proceed as follows:

$x^2-x+1=0hArrx^2-2xx x xx (1/2)+(1/2)^2-(1/2)^2+1=0$

or $(x-1/2)^2+3/4=0$

i.e. $(x-1/2)^2-(sqrt3/2i)^2=0$

i.e. $(x-1/2+sqrt3/2i)(x-1/2-sqrt3/2i)=0$

i.e. $x=1/2-sqrt3/2i$ or $x=1/2+sqrt3/2i$

Hence, three cube roots of $1$ are ${-1,1/2-sqrt3/2i,1/2+sqrt3/2i}$

How do you find three cube roots of -1-1?