What are the three cube roots of -1? Not sure if

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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?

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Expert2022-01-19Added 777 answers

Step 1 Let xx in 1x31x=1+x+x2 Generally suppose f(x) is a polynomial over a field with roots ab. Then f(x)=(xa)g(x) hence f(b)=0 (ab)g(b)=0 g(b)=0 i.e. b is a root of f(x)(xa) Step 2 From a factorization perspective, the reason that this works is because, over a domain, monic linear polynomials are ,


Expert2022-01-19Added 669 answers

Write -1 in polar form as eiπ.
In general, the cube roots of
are given by
r13eiθ3, r13ei(θ3+2π3)
In your case r=1 and θ=π, so your cube roots are eiπ3eiπ, and ei5π3 Put back into rectangular form, they are
12+i32, 1 and 12i32

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