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???
Applying simple substitutions: $\sqrt[3]{a}=m, \sqrt[3]{b}=n, \sqrt[3]{c}=k$, we have
$$(mnk)^3=1, m+n+k=0 \\ mnk=1, m+n+k=0 \\ m^3+n^3+k^3=m^3+n^3-(m+n)^3 =-3mn(m+n)=\frac{-3}{k} \times (-k)=3.$$
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 $$a+b+c-3\sqrt[3]{abc}=0\implies a+b+c=3.$$
Use the relationship,
$$ x^3+y^3+z^3 = (x+y+z)^3 - 3(x+y)(y+z)(z+x)$$
to evaluate,
$$a+b+c= (\sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{c})^3 -3(\sqrt[3]{a} + \sqrt[3]{b})(\sqrt[3]{b} + \sqrt[3]{c})(\sqrt[3]{c} + \sqrt[3]{a} )$$
$$= 0^3 -3(0- \sqrt[3]{c})(0- \sqrt[3]{a})(0- \sqrt[3]{b} )=3\sqrt[3]{abc}=3$$
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:
$a + b + c = -3 \left(\sqrt[3]{a} \cdot \sqrt[3]{b} \right) \left(\sqrt[3]{a} + \sqrt[3]{b} \right) = -3 c^{-\frac{1}{3}} (-c^{\frac{1}{3}}) = 3$.
$$ \begin{aligned} a+b+c-3\sqrt[3]{abc}&=\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\left(\sqrt[3]{a^{2}}+\sqrt[3]{b^{2}}+\sqrt[3]{c^{2}}-\sqrt[3]{ab}-\sqrt[3]{ac}-\sqrt[3]{bc}\right)\\ \\ &=0 \end{aligned} $$
