On p. 113, I wrote, "I'd have to introduce an unreasonable amount of extra mathematics in order to explain why one [gluon] is extra." Briefly, the mathematics in question is the decomposition $U(3) = SU(3) \times U(1)$. The way $U(3)$ arises is that the endpoint of a string ending on any of three coincident D3-branes carries on it a Hilbert space spanned by three basis vectors, call them $|R\rangle$, $|G\rangle$, and $|B\rangle$. The group $U(3)$ arises as the isometries of this Hilbert space. Its $U(1)$ center rotates the phase of each of these basis vectors equally, and amounts to uniform translation of all three branes by a fixed distance. This is the Goldstone mode of the branes. Because of supersymmetry, the Goldstone mode (a scalar on the brane worldvolume) is accompanied by fermions and a gauge boson. This gauge boson is the "extra" gluon---except it's not a gluon at all, because it doesn't interact with the $SU(3)$ gluons.
Just to get the counting straight: $U(3)$ has nine generators. $U(1)$ has one. To get the $8$ gluons that I mentioned in the text of the book, I just say $9-1=8$.
Incidentally, when I later say (on p. 114) that there are about $N^2$ gluons when we put $N$ D3-branes together, I basically have the same calculation in mind: The Hilbert space on the end of the string is now $N$-dimensional, there's a $U(N)$ symmetry, and the interacting gauge bosons correspond to the $SU(N)$ subgroup, which has dimension $N^2-1$.