Tensor Property Show
$$
(Q^3)_{ij}-\frac{1}{2}Tr(Q^2)Q_{ij}-\det(Q)\delta_{ij}=0
$$
where Q is a real symmetric traceless tensor. $δ_{ij}$ is Kronecker delta symbol which is 1 if $i=j$ or $0$ otherwise. We can possibly use eigenvalues & eigenvectors of Q to help us with this, I am not sure how to go about it though. Also given is
$$
Q_{ij}=\epsilon_{ij}-\frac{1}{3}Tr(\epsilon)\delta_{ij}$$
where $\epsilon$ is the totally anti-symmetric tensor-Levi-Civita symbol.  This post recently got taken off because it lacked explanation. This is a math problem from Russian math books by Demidovich, not sure what the problem is, please do not delete the post if you cannot solve it.  It is just a tensor property that we should all know, I am trying :)
 A: Since $Q$ is a $3\times3$ matrix, Cayley-Hamilton theorem says that $Q^3-tQ^2+sQ-rI=0$, for some specific scalar $(r,s,t)$. For example, $t$ is the trace of $Q$, hence $t=0$ in the present case, and $r$ is the determinant of $Q$. Hence the identity to prove holds as soon as 
$$s=-\tfrac12\mathrm{tr}(Q^2).
$$
The characteristic polynomial $x^3-tx^2+sx-r$ is the determinant of $xI-Q$ hence $s$ is the derivative at $x=0$ of this determinant, that is,
$$
s=\sum_i\det(Q^{(i,i)}),
$$
where $Q^{(i,i)}$ is the $2\times2$ matrix obtained by cancelling the $i$th column and the $i$th row of $Q$. Thus,
$$
s=\tfrac12\sum_{i\ne j}(Q_{ii}Q_{jj}-Q_{ij}Q_{ji}).
$$
Furthermore, the trace $t$ is zero hence
$$
t^2=0=\sum_iQ_{ii}^2+\sum_{i\ne j}Q_{ii}Q_{jj}.
$$
Finally,
$$
\mathrm{tr}(Q^2)=\sum_i(Q^2)_{ii}=\sum_{i,j}Q_{ij}Q_{ji}=\sum_{i}Q_{ii}^2+\sum_{i\ne j}Q_{ij}Q_{ji}.
$$
Massaging the three last displayed identities above yields the desired one.
All this only uses that $Q$ is a $3\times 3$ traceless matrix.
