Doubt in finding general term of the given sequence

The author is assuming the original terms of the series are values from a polynomial at increasing integral values. Thus, when a sequence of differences causes all of the terms to be the same at some level $m$, the sequence $T_n$ represents a polynomial of degree $m$. For more information, see Theorem $1$ in Difference Tables of Sequences. In your case, taking the sequence of the third order differences based on those second order differences causes all of the values to be $6$. This means the original sequence represents a cubic equation.

Also, although the author doesn't use this, and the linked page only hints at it, the sequence of constant values are $a_n(n!)$ where $a_n$ is the coefficient of the $x^n$ term. Thus, in this case, $6 = a_3(3!) \implies a_3 = 1$. The author could have used this to determine that $a = 1$ more directly.