= Solution
The <Cartesian tensor> law is $T'_{ik}=Q_{ia}Q_{kb}T_{ab}$. In every complete <tensor contraction>, summation over a transformed index contracts a pair of $Q$ factors to a <Kronecker delta>. In detail,
$$
\alpha'=Q_{ia}Q_{ib}T_{ab}=\delta_{ab}T_{ab}=\alpha,
$$
$$
\beta'=Q_{ia}Q_{kb}Q_{kc}Q_{id}T_{ab}T_{cd}=\delta_{bc}\delta_{ad}T_{ab}T_{cd}=T_{ab}T_{ba}=\beta,
$$
$$
\gamma'=Q_{ia}Q_{kb}Q_{kc}Q_{md}Q_{me}Q_{if}T_{ab}T_{cd}T_{ef}=\delta_{bc}\delta_{de}\delta_{af}T_{ab}T_{cd}T_{ef}=T_{ab}T_{bd}T_{da}=\gamma.
$$
Thus all three are invariant under <rotations>. They are $\operatorname{tr}T$, $\operatorname{tr}(T^2)$ and $\operatorname{tr}(T^3)$. For a real <symmetric tensor>, <orthogonal diagonalization> gives
$$
\alpha=\sum_i\lambda_i,\qquad\beta=\sum_i\lambda_i^2,\qquad\gamma=\sum_i\lambda_i^3.
$$
Expanding $\prod_i(\lambda-\lambda_i)$ and using <Newton's identities> to recover the <characteristic polynomial from traces of powers> yields
$$
\boxed{\lambda^3-\alpha\lambda^2+\frac{\alpha^2-\beta}{2}\lambda-\frac{\alpha^3-3\alpha\beta+2\gamma}{6}=0.}
$$
Comparison identifies $\boxed{c_2=-1,\qquad c_1=\tfrac12,\qquad c_0=-\tfrac16}$.
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