Easy exponents

From Newton's sums, we know that

$A^3+B^3+C^3 = (A+B+C)(A^2+B^2+C^2) - (AB+AC+BC)(A+B+C) + 3ABC$

Given that $A+B=-C \implies A+B+C=0$

$\dfrac{1}{3}A^3 +\dfrac{1}{3}B^3 + \dfrac{1}{3}C^3\\ =\dfrac{1}{3}(A^3+B^3+C^3)\\ =\dfrac{1}{3}\left((A+B+C)(A^2+B^2+C^2) - (AB+AC+BC)(A+B+C) + 3ABC\right)\\ =\dfrac{1}{3}\left((0)(A^2+B^2+C^2) - (AB+AC+BC)(0) + 3ABC\right)\\ =\dfrac{1}{3}(3ABC)\\ =\boxed{ABC}$

(A+B)^{3}=-C^{3}

A^{3}+B^{3}+3AB(A+B)+C^{3}=0

A^{3}+B^{3}+C^{3}+3AB(-C)=0

A^{3}+B^{3}+C^{3}-3ABC=0

A^{3}+B^{3}+C^{3}=3ABC

$\frac{1}{3}$ A^{3}+ $\frac{1}{3}$ B^{3}+ $\frac{1}{3}$ C^{3}= $\frac{1}{3}$ 3ABC

$\frac{1}{3}$ A^{3}+ $\frac{1}{3}$ B^{3}+ $\frac{1}{3}$ C^{3}=ABC