Derive the Simpson's $\frac{3}{8}$ rule
Simpson's $\frac{3}{8}$ rule for integration can be derived by approximating the given function $f(x)$ with the $3^{\text{rd}}$ order(cubic) polynomial $f_3(x)$ $$f_3(x)=a_0+a_1x+a_2x^2+a_3x^3$$ Using Lagrange interpolation, the cubic polynomial function $f_3(x)$ that passes through $4$ can be explicitly given as $$f_3(x)=\frac{(x-x_1)(x-x_2)(x-x_3)}{(x_0-x_1)(x_0-x_2)(x_0-x_3)}f(x_0)+\frac{(x-x_0)(x-x_2)(x-x_3)}{(x_1-x_0)(x_1-x_2)(x_1-x_3)}f(x_1)+\frac{(x-x_0)(x-x_1)(x-x_3)}{(x_2-x_0)(x_2-x_1)(x_2-x_3)}f(x_2)+\frac{(x-x_0)(x-x_1)(x-x_2)}{(x_3-x_0)(x_3-x_1)(x_3-x_2)}f(x_3)$$ \begin{align} I&=\int_a^bf(x)\:dx\\ &\approx \int_a^bf_3(x)\:dx \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad(1)\\ &=(b-a)\times \frac{f(x_0)+3f(x_1)+3f(x_2)+f(x_3)}{8} \end{align}
Whenever I tried to integrate $(1)$ I completely lost. My Lecturer didn't provide any other method even all previous derivation$(\text{Simpson's }\frac{1}{3},\text{trapezoid})$ was also done by this. Lagrange method is mostly a theoretical tool used for proving those theorems. Not only it is not very efficient when a new point is added (which requires computing the polynomial again, from scratch). I feel there must be another way to derive those formula by easily and I got this. But which isn't familiar with me and I failed to apply it in those derivation.
I heartily thank if anyone explain the linked rule in details.