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???

Use the relationship,
$$ x^3+y^3+z^3 = (x+y+z)^3 - 3(x+y)(y+z)(z+x)$$
to evaluate,
$$a+b+c= (\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c})^3
-3(\sqrt[3]{a} + \sqrt[3]{b})(\sqrt[3]{b} + \sqrt[3]{c})(\sqrt[3]{c} + \sqrt[3]{a} )$$
$$= 0^3 -3(0- \sqrt[3]{c})(0- \sqrt[3]{a})(0- \sqrt[3]{b} )=3\sqrt[3]{abc}=3$$
