<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] Original problem : $$\color{green}{(a-b)}\color{red}{(2b-a+2)}\color{green}{(2b+1)} - (b+1)\color{blue}{(2a-2b-1)}(2b-a)$$

*

*) Notice how

$\color{green}{(a-b)(2b+1)}=-2b^2+2ab-b+a=b(-2b+2a-1)+a=b\color{blue}{(2a-2b-1)}+a$
$$\Rightarrow\color{blue}{(2a-2b-1)}=\dfrac{\color{green}{(a-b)(2b+1)}-a}{b}$$


*) We find the same for the second part of the equation:

$(b+1)(2b-a)=2b^2-ab+2b-a=b\color{red}{(2b-a+2)}-a$
$$\Rightarrow\color{red}{(2b-a+2)}=\dfrac{(b+1)(2b-a)+a}{b}$$
Plugging in we get:
$$(a-b)\color{red}{\dfrac{(b+1)(2b-a)+a}{b}}(2b+1) - (b+1)(2a-2b-1)(2b-a)=$$
$$\dfrac{(a-b)}{b}\cdot[(b+1)(2b-a)+a]\cdot(2b+1) - (b+1)(2a-2b-1)(2b-a)=$$
$$\dfrac{(a-b)}{b}\cdot(2b+1)\cdot[(b+1)(2b-a)+a] - (b+1)(2a-2b-1)(2b-a)=$$
$$\dfrac{(a-b)}{b}\cdot(2b+1)\cdot a + \dfrac{(a-b)}{b}(2b+1)(b+1)(2b-a) - (b+1)(2a-2b-1)(2b-a)=$$
$$\dfrac{(a-b)}{b}(2b+1)a + (b+1)(2b-a)\left(\dfrac{(a-b)}{b}(2b+1)+(2a-2b-1)\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{(a-b)}{b}(2b+1)+(2a-2b-1)\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}+\dfrac{\color{green}{(a-b)(2b+1)}-a}{b}\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{b\color{blue}{(2a-2b-1)}+\color{green}{(a-b)(2b+1)}}{b}\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{(2ab-2b^2-b)+\color{green}{-2b^2+2ab-b+a}}{b}\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + (b+1)(2b-a)\left(\dfrac{4ab-4b^2-2b+a}{b}\right)$$
$$\dfrac{b\color{blue}{(2a-2b-1)}+a}{b}a + \dfrac{b\color{red}{(2b-a+2)}-a}{b}\left({4ab-4b^2-2b+a}\right)$$
$$\color{blue}{(2a-2b-1)}a+\dfrac{a^2}{b} + \left(\color{red}{(2b-a+2)}-\dfrac{a}{b}\right)\left({4ab-4b^2-2b+a}\right)$$
There is definitely something in this 'equation', but it definitely won't save you either time or paper.
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