= Solution
The <tensor contraction> $C_{mkl}=\epsilon_{ijm}T_{ijkl}$ is an <isotropic third-rank tensor> under proper <rotations>. It therefore equals $\nu\epsilon_{mkl}$. Cyclically permuting three indices is even, so $\epsilon_{mkl}=\epsilon_{klm}$ and \b[$\epsilon_{ijm}T_{ijkl}=\nu\epsilon_{klm}$].
For the given <Kronecker delta> expression, the $\alpha$ term vanishes by antisymmetry, while the $\beta$ and $\gamma$ terms give respectively $\beta\epsilon_{klm}$ and $\gamma\epsilon_{lkm}=-\gamma\epsilon_{klm}$. Hence
$$
\boxed{\epsilon_{ijm}T_{ijkl}=(\beta-\gamma)\epsilon_{klm},\qquad \nu=\beta-\gamma.}
$$
These computations also illustrate <isotropic contractions of a fourth-rank tensor>.
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