2022-01-17

What are the three cube roots of -1?
Not sure if this is a trick question, But I have been asked this. one of the ansers is -1, what are the other 2?

nick1337

Step 1 Let $x\to -x$ in $\frac{1-{x}^{3}}{1-x}=1+x+{x}^{2}$ Generally suppose f(x) is a polynomial over a field with roots $a\ne b$. Then $f\left(x\right)=\left(x-a\right)g\left(x\right)$ hence $f\left(b\right)=0$ $⇒\left(a-b\right)g\left(b\right)=0$ $g\left(b\right)=0$ i.e. b is a root of $\frac{f\left(x\right)}{\left(x-a\right)}$ Step 2 From a factorization perspective, the reason that this works is because, over a domain, monic linear polynomials are ,

Vasquez

Write -1 in polar form as ${e}^{i\pi }$.
In general, the cube roots of
$r{e}^{i\theta }$
are given by

${r}^{\frac{1}{3}}{e}^{i\left(\frac{\theta }{3}+\frac{4\pi }{3}}.$
In your case $r=1$ and $\theta =\pi$, so your cube roots are ${e}^{\frac{i\pi }{3}}{e}^{i\pi }$, and ${e}^{i}5\frac{\pi }{3}$ Put back into rectangular form, they are
and $\frac{1}{2}-i\frac{3}{2}$

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