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?

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

Expert2022-01-19Added 777 answers

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 $f(x)=(x-a)g(x)$ hence
$f(b)=0$
$\Rightarrow (a-b)g(b)=0$
$g(b)=0$
i.e. b is a root of
$\frac{f(x)}{(x-a)}$
Step 2
From a factorization perspective, the reason that this works is because, over a domain, monic linear polynomials are ,

Vasquez

Expert2022-01-19Added 669 answers

Write -1 in polar form as

In general, the cube roots of

are given by

In your case

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