Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$ Let $a,$ $b,$ $c$ be real numbers such that $abc = 1$ and $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0.$ Find $a + b + c.$

I don't really know how to approach this. I was thinking doing something like squaring or cubing $\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c} = 0$ would help, but it doesn't really work out...Any help???

Note that $abc = 1 \implies \sqrt[3]{a} \cdot \sqrt[3]{b} = c^{-\frac{1}{3}}$.
Now, cube both sides of $\sqrt[3]{a} + \sqrt[3]{b} = - \sqrt[3]{c}$ and re-arrange the terms to get:
$a + b + c = -3 \left(\sqrt[3]{a} \cdot \sqrt[3]{b} \right) \left(\sqrt[3]{a} + \sqrt[3]{b} \right) = -3 c^{-\frac{1}{3}} (-c^{\frac{1}{3}}) = 3$. 
