if a,b,c are roots of a cubic equation then for the following question... If $a, b, c$ are roots of $x^3 -3x^2 + 2x +4 = 0$ and
$$y= 1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}$$
then value of $y$ at $x=2$ is:

$(x-a)(x-b)(x-c)+a(x-b)(x-c)+bx(x-c)+cx^2=x^3,$
which says that
$$y=\frac{x^3}{(x-a)(x-b)(x-c)}.$$
Thus, $$y(2)=\frac{8}{(2-a)(2-b)(2-c)}=\frac{8}{8-4\cdot3+2\cdot2-(-4)}=2.$$
