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???

There's a remarkable identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx).$$
Applying this to $x=\sqrt[3]a,y=\sqrt[3]b,z=\sqrt[3]c$ gives that
$$a+b+c-3\sqrt[3]{abc}=0\implies a+b+c=3.$$
