Good night!... again.

Let $x = 3 - 2a$ , $y = 3 - 2b$ , and $z = 3 - 2c$ . Then $a - b = \frac{1}{2}(y - x)$ , $a - c = \frac{1}{2}(z - x)$ , and $b - c = \frac{1}{2}(z - y)$ . Substituting these values into the expression gives

$\frac{xy}{-\frac{1}{2}(z - x) \cdot -\frac{1}{2}(z - y)} + \frac{yz}{\frac{1}{2}(y - x) \cdot \frac{1}{2}(z - x)} + \frac{zx}{\frac{1}{2}(z - y) \cdot -\frac{1}{2}(y - x)}$

which simplifies to

$4\bigg[\frac{xy}{(z - x)(z - y)} + \frac{yz}{(x - y)(x - z)} + \frac{xz}{(y - z)(y - x)}\bigg]$ .

Using the identity $\frac{xy}{(z - x)(z - y)} + \frac{yz}{(x - y)(x - z)} + \frac{xz}{(y - z)(y - x)} = 1$ , the expression further simplifies to $\boxed{4}$ .

Using Lagrange-Interpolation . $\dfrac{(3 - 2a)(3 - 2b)}{(c - a)(c - b)} + \dfrac{(3 - 2b)(3 - 2c)}{(a - b)(a - c)} + \dfrac{(3 - 2c)(3 - 2a)}{(b - c)(b - a)}$ $=4\times\dfrac{(1.5 - a)(1.5- b)}{(c - a)(c - b)} +4\times \dfrac{(1.5 - b)(1.5 - c)}{(a - b)(a - c)} +4\times \dfrac{(1.5- c)(1.5- a)}{(b - c)(b - a)}$ Let $f(x)=4\times\dfrac{(x - a)(x- b)}{(c - a)(c - b)} +4\times \dfrac{(x - b)(x - c)}{(a - b)(a - c)} +4\times \dfrac{(x- c)(x- a)}{(b - c)(b - a)}$ Then $f(x)$ is the lowest degree polynomial that passes through the points $(a,4),(b,4),(c,4)$ ,so $f(x)=4,f(1.5)=4$

The expression is as follows:

$\begin{aligned} S & = \sum_{cyc} \frac {(3-2a)(3-2b)}{(c-a)(c-b)} \\ & = \sum_{cyc} \frac {4\left(\frac 32-a\right)\left(\frac 32-b\right)}{(c-a)(c-b)} & \small \color{#3D99F6} \text{Let }x=\frac 32 - a, \ y=\frac 32 - b, \ z=\frac 32 - c \\ & = \sum_{cyc} \frac {4xy}{(x-z)(y-z)} \\ & = 4 \sum_{cyc} \frac {xy(y-x)}{(x-y)(y-z)(z-x)} \\ & = 4\times \frac {xy(y-x)+yz(z-y)+zx(x-z)}{(x-y)(y-z)(z-x)} \\ & = 4 \times \frac {xy^2-x^2y+yz^2-y^2z+zx^2-z^2x}{xy^2-x^2y+yz^2-y^2z+zx^2-z^2x} \\ & = \boxed{4} \end{aligned}$

You should know these:

$\dfrac{1}{(c - a)(c - b)} + \dfrac{1}{(a - b)(a - c)} + \dfrac{1}{(b - c)(b - a)} = 0$

$\dfrac{a + b}{(c - a)(c - b)} + \dfrac{b + c}{(a - b)(a - c)} + \dfrac{c + a}{(b - c)(b - a)} = 0$

$\dfrac{ab}{(c - a)(c - b)} + \dfrac{bc}{(a - b)(a - c)} + \dfrac{ca}{(b - c)(b - a)} = 1$

People are probably going to write a messy solution using cyclic sum and semplification. The problem has a numeric solution, so, as long as the denominators are not 0, one can put whatever numbers he want and still obtain the identity. Putting a=1, b=2, c=3, you get the solution, 4.