<s>[INST] 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???
 [/INST] 
Applying simple substitutions: $\sqrt[3]{a}=m, \sqrt[3]{b}=n, \sqrt[3]{c}=k$, we have
$$(mnk)^3=1, m+n+k=0 \\ mnk=1, m+n+k=0 \\ m^3+n^3+k^3=m^3+n^3-(m+n)^3 =-3mn(m+n)=\frac{-3}{k} \times (-k)=3.$$

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