What are the three cube roots of -1?

Question

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?

Answer 1

Write $-1$ in polar form as $e^{i\pi}$. In general, the cube roots of $r e^{i\theta}$ are given by $r^{1/3}e^{i\theta/3}$, $r^{1/3}e^{i(\theta/3 + 2\pi /3)}$ and $r^{1/3}e^{i(\theta/3 + 4\pi /3)}$. In your case $r = 1$ and $\theta = \pi$, so your cube roots are $e^{i\pi / 3}$, $e^{i\pi}$, and $e^{i 5\pi/ 3}$. Put back into rectangular form, they are ${1 \over 2} + i{\sqrt{3} \over 2}$, $-1$, and ${1 \over 2} - i{\sqrt{3} \over 2}$.

Answer 2

HINT $\ $ Let $\rm\ \ x\ \to\ -x\ \ $ in $\rm\displaystyle\ \ \frac{1-x^3}{1-x}\ =\ 1+x+x^2\:.\ $

Generally suppose $\rm\:f(x)\:$ is a polynomial over a field with roots $\rm\: a \ne b\:$. Then $\rm\ f(x) = (x-a)\ g(x)\ $ hence $\rm\: f(b) = 0\: \Rightarrow\ (a-b)\:g(b) = 0\ \Rightarrow\ g(b) = 0\ $ i.e. $\rm\:b\:$ is a root of $\rm\ f(x)/(x-a)\:$.

From a factorization perspective, the reason that this works is because, over a domain, monic linear polynomials are prime, so the linear factors of a polynomial are unique, i.e. the roots and their multiplicity are unique. e.g. see my post here. This fails over coefficient rings that are not domains, i.e. have zero-divisors, e.g. $\rm\ x^2-1 = (x-1)(x+1) = (x-4)(x+4)\ $ over $\ \mathbb Z/15\:$. Here, although $4 \ne 1$ is a root of $\rm\ x^2 - 1$ it is not true that 4 is a root of $\rm\ (x^2-1)/(x-1) = x+1\:$. For the example at hand we have $\rm\ x^3 + 1 = (x+1)(x+9)(x-10) = (x+16)(x+22)(x-38)\ $ over $\ \mathbb Z/91\:$.

Answer 3

You wish to solve for $z$ in $z^3=-1$. You can put

\begin{align} -1&=e^{\pi i}\\ &=e^{\pi i + 2\pi i k}\\ &=e^{\pi i (2k+1)} \end{align}

Thus the solutions for $z$ are

$$e^{\frac{\pi i}{3}(2k+1)}$$

for $k=0,1,2$, so we have

$$e^{\frac{\pi i}{3}},-1, e^{\frac{5\pi i}{3}}$$

as the $3$ cubic roots of $-1$.

Answer 4

You may take a look at this previous question.