How is it possible to rewrite the matrix product as a commutator in a Lie algebra? so suposing $(t^m)_{m \in M}$ are the generators of a Lie group, and we have the following relation in the Lie algebra:
$$[t^a,t^b]= i f^{abc}t^c$$
Where the $f^{abc}$ are the structure constants of the algebra.
And in a irreducible representation ($r$):   $  \,\,\,\,\,\,t_r^at_r^a = C_2(r) I$
where $C_2(r)$ is the the quadratic Casimir operator of the representation $r$.
In a book of QFT the following equalities are written but I can't seem to understand why they are correct:
$$t^bt^at^b= t^bt^bt^a+t^b[t^a,t^b] \,\,\,\,\,\,\,\, \text{this one really is obvious}$$
$$\;\;\;\;\;\;=C_2(r)t^a + i t^bf^{abc}t^c\,\,\, \text{this one is also obvious}$$
$$\hspace{3.1cm}=C_2(r)t^a + \frac{1}{2}if^{abc}if^{bcd}t^d \,\,\, \text{??? what just happened here?}$$
In the text it says that "In the third line we have used the antisymmetry of $f^{abc}$ to rewrite the matrix product as a commutator;" and I can't see how that would be possible.
One cannot write $t^bt^c$ as a simple sum of commutators, if we were to try:
$t^bt^c-t^ct^b+t^ct^b-t^bt^c = [t^b,t^c] + [t^c,t^b]=0$
$t^bt^c-t^ct^b+t^ct^b+t^bt^c = [t^b,t^c] + \{t^b,t^c\} = 2t^bt^c $
So only if $\{t^b,t^c\}=0$ we would get that result, but I can't see why would that be.

Your QFT book would be remiss if it did not stress that it has chosen its Lie algebra basis s.t. the structure constants  are antisymmetric in all three indices, not just a,b.
From this antisymmetry in b,c as well, you see that the bilinear in generators is antisymmetrized, and the rest follows trivially. The anticommutator does not vanish: it simply does not enter in the expression.
