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
 A: This is not really a general method but just an insight. I would look at the terms and see if they can be written as sums of common terms. In your example:
$$2b+1 = b + (b+1)$$
$$2b-a = b - (a-b)$$
$$2b-a+2 = (b+1) + (-a+b+1)$$
$$2a-2b-1 = (a-b) - (-a+b+1)$$
Using the above, the expression can be re-written as:
$$(a-b)[(b+1)+(-a+b+1)][b+(b+1)] \; - \; (b+1)[(a-b)-(-a+b+1)][b-(a-b)]$$
At this point it seems that the expression can be simplified, because it involves some products of same terms with opposite signs that cancel each other out. If you like something tidier, you can take $X=(a-b)$ , $Y=(b+1)$ , $Z=(-a+b+1)$ , and $T=b$ . Then the above equation becomes:
$$X(Y+Z)(Y+T) + Y(X-Z)(X-T) $$
which makes it more intuitive to continue:
$$X(Y^2 + (Z+T)Y + ZT) \; + \; Y(X^2 -(Z+T)X +ZT)$$
$$= \; (X+Y)(XY + ZT)$$
Now getting back to the definitions of our auxiliary variables we see that
$$X+Y=a+1$$
$$XY + ZT = (a-b)(b+1) + (-(a-b)+1)(b) = (a-b)b + a-b -(a-b)b + b = a$$
So the original expression simplifies to $(a+1)a$ .
A: 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.
A: 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!
A: Let the left side be $f(a,b),$ a polynomial in $a$ and $b.$  It is easily confirmed that  the coefficient of $b^3$ is zero. Thus, $$f(a,b)=c_0(a)+c_1(a)b+c_2(a)b^2.$$ It is easy to check that
$$ f(a,0)=f(a,a)=f(a,-1)=a^2+a.$$
Thus, by interpolation (considering $f(a,b)$ as a polynomial of degree two in $b$),
we conclude that $f(a,b)=a^2+a.$
