Find the value of $(-1)^{1/3}$.

Question

Evaluate $(-1)^{\frac{1}{3}}$.

I've tried to answer it by letting it be $x$ so that $x^3+1=0$. But by this way, I'll get $3$ roots, how do I get the actual answer of $(-1)^{\frac{1}{3}}$??

Answer 1

It depends whether you are trying to solve it as an equation over the reals ($\mathbb{R}$) or over the complex plane ($\mathbb{C}$).
The polynomial $x^3+1$ factors as $x^3+1=(x+1)(x^2-x+1)$. So in the first case, you have exactly one solution: $x=-1$, since the second polynomial has no real roots. If you're looking for roots over $\mathbb{C}$, then you'll have three roots, since the $\sqrt[3]t$ is not a function over $\mathbb{C}$, hence all the 3 roots have equal rights to be called "the root".

Answer 2

Have you tried looking for factors of $(x+y)^3$ ??

$(x+y)^3 = (x+y)(x^2-xy+y^2)$ therefore $(x+1)^3 = (x+1)(x^2-x+1) \Rightarrow x=-1$ is one of the root, and apply methods you learnt to find roots of quadratic equation to find other roots.

Answer 3

They are all equally considered to be values of $\sqrt[3]{-1}$. There is no unique cube root, just as there is no unique square root.

Answer 4

Just put it like complex numbers:

We know that $z=\sqrt[k]{m_\theta}$, so

$z=\sqrt[3]{-1}$

$-1=1_{\pi}$

$\alpha_n=\dfrac{\theta+k\pi}{n}$

$\alpha_0=60$

$\alpha_1=180$

$\alpha_2=300$

So the answers are:

$z_1=1_{\pi/3}$

$z_2=1_{\pi}$

$z_3=1_{5\pi/3}$