Triple Cube Sum

We start by simplifying the second and third equations. Simplifying the second equation gives us $\frac{a}{b} + \frac{c}{a} + \frac{b}{c} = -1$ Then, we add 3 to both sides of the third equation to obtain $\frac{a^2}{bc} + \frac{b^2}{ac} + \frac{c^2}{ab} + \frac{bc}{a^2} + \frac{ac}{b^2} + \frac{ab}{c^2} + 3 = 1$ which factors as follows: $\left(\frac{a}{b} + \frac{c}{a} + \frac{b}{c}\right) \left(\frac{b}{a} + \frac{a}{c} + \frac{c}{b}\right) = 1$ Thus, by combining both results above, we conclude that $\frac{b}{a} + \frac{a}{c} + \frac{c}{b} = -1$ Then, we have that $\left(\frac{a}{b} + \frac{c}{a} + \frac{b}{c}\right) + \left(\frac{b}{a} + \frac{a}{c} + \frac{c}{b}\right) = \frac{a^2 b + a^2 c + a b^2 + a c^2 + b^2 c + b c^2}{abc} = -2$ or, rewritten slightly, $a^2 b + a^2 c + a b^2 + a c^2 + b^2 c + b c^2 + 2abc = (a + b)(b + c)(a + c) = 0$ Finally, we note the identity $\begin{aligned} (a + b + c)^3 & = a^3 + b^3 + c^3 + 3(a + b)(b + c)(a + c) \\ & = a^3 + b^3 + c^3 \text{ (by the previous result)} \end{aligned}$ Hence, $a^3 + b^3 + c^3 = (a + b + c)^3 = 6^3 = \boxed{216}$ .