Find the value of $(-1)^{1/3}$. 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 $(-1)^{\frac{1}{3}}$?? 
 A: 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".
A: 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.
A: 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.
A: 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:
$z_1=1_{\pi/3}$
$z_2=1_{\pi}$
$z_3=1_{5\pi/3}$
