How to evaluate $\int_a^b(b-x)(x-a)dx$ in a smart way? It holds that $\int_a^b(b-x)(x-a)dx=1/6(b-a)^3$. I was wondering if there is a smart way of seeing this, instead of brute-force solving the intergral (it's not that tedious, but if there is a quick way about it, I would like to know it)
If it's not possible to do it smartly, then my other question would be how we go about factorizing the following expression:
$$
1/6(b^3-a^3+3a^2b-3ab^2).
$$
I mean, I know now that it equals $1/6(b-a)^3$, but what are the steps for factorizing this? Should we think of it as a polynomial in $a$ and $b$? Whenever I think of a factorization of a polynomial, I think of zeroes, but I'm not sure how that would work here.

Let $u=x-a$. Then $b-x=b-a-u$.
$$\int_a^b(b-x)(x-a)dx=\int_0^{b-a}[(b-a)u-u^2]du=\left[\frac{(b-a)u^2}{2}-\frac{u^3}{3}\right]_0^{b-a}$$
