The generators of $SU(N)$ have the following properties: $Trace[T^{a}T^{b}]=\frac{1}{2} \delta^{ab}---(i)$ and $Trace[T^{a}]=0---(ii)$. In product form $$T^{a}T^{b}=\frac{1}{2}(\frac{1}{N} \delta^{ab} I+(d_{abc}+if_{abc})T^{c})$$ Note that sum is implied over repeated index
Where $d_{abc}=2Trace[ \{ T^{a},T^{b} \} T^{c}]$ and $f_{abc}=2iTrace[ [T^{a},T^{b}]T^{c}]$ I want to simplify the expression $Trace[T^{a}T^{b}(T^{N^{2}-1})^{2}]$. First, I use the product identity on $T^{a}T^{b}$: $$Trace[T^{a}T^{b}(T^{N^{2}-1})^{2}]=Trace[\frac{1}{2}(\frac{1}{N} \delta^{ab} I+(d_{abc}+if_{abc})T^{c})(T^{N^{2}-1})^{2}]$$ $$=\frac{1}{2}Trace[\frac{1}{N} \delta^{ab} (T^{N^{2}-1})^{2}+(d_{abc}+if_{abc})T^{c}(T^{N^{2}-1})^{2}]$$ Using $(i)$ and apply the the product formula in the second term again, it simplifies to: $$\frac{1}{4N} \delta^{ab}+\frac{1}{4}Trace[(d_{abc}+if_{abc})(\frac{1}{N} \delta^{c,N^{2}-1} I+(d_{c,N^{2}-1,e}+if_{c,N^{2}-1,e})T^{e})T^{N^{2}-1}]$$ Applying $(i)$ and $(ii)$ and the fact that $f{abc}$ is totally antisymmetric, it gives: $$\frac{1}{4N} \delta^{ab}+\frac{1}{8}(d_{abc}+if_{abc})d_{c,N^{2}-1,N^{2}-1}$$ Can the sum over $c$ in this expression be further simplified?