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For a finite-dimensional complex representation $V$, the character of its exterior cube isWrite $p_r={\sum }_i{\lambda }_i^r={\chi }_V(g^r)$ for the power sums of the eigenvalues of $g$. The trace on the exterior cube is the elementary symmetric polynomial $e_3={\sum }_{i<j<k}{\lambda }_i{\lambda }_j{\lambda }_k$. Newton's identities give $6e_3=p_1^3-3p_1{p}_2+2p_3$, proving the formula.

For a $d$-dimensional vector space, evaluation at the identity gives $(d^3-3d^2+2d)/6=\left(\genfrac{}{}{0ex}{}{d}{3}\right)$. The formula therefore vanishes for $d<3$, as the exterior cube must.