<s>[INST] Algebraic simplification of $(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)$ When trying to replicate a proof about Catalan numbers, I came across the following simplification $$(a-b)(2b-a+2)(2b+1) - (b+1)(2a-2b-1)(2b-a)= a^2+a$$
This is confirmed by writing out all the terms (and by WolframAlpha). My question - is there a clever substitution or factorisation that makes this obvious / less tedious? (Or can anyone provide a more succinct explanation of the answer linked above?)
Thanks
 [/INST] I found a solution from Lord Commander's working. As they note:
$$(a-b)(2b+1) = b(2a-2b-1) + a \tag{1}\label{eq1}$$
And:
$$(b+1)(2b-a) = b(2b-a+2) -a \tag{2}\label{eq2}$$
For simplicity, let $x = 2a-2b-1$ and $y = 2b-a+2$.
Thus $\eqref{eq1}$ becomes $(a-b)(2b+1) = bx + a$ and $\eqref{eq2}$ becomes $(b+1)(2b-a) = by - a$
Applying the equations:
\begin{align} (a-b)(2b+1)(2b-a+2) - (b+1)(2b-a)(2a-2b-1) &= y(bx + a) - x(by-a) \\ 
&= bxy + ay -bxy + ax \\
&= a(x+y) \\
&= a(2a-2b-1+2b-a+2) \\
&= a(a+1) \\
&= a^2 + a 
\end{align}
Thanks Lord Commander!
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