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:
 A: $$1+\dfrac a{x-a}=\dfrac x{x-a}$$
$$1+\dfrac a{x-a}+\dfrac{bx}{(x-a)(x-b)} =x\left[\dfrac1{x-a}+\dfrac b{(x-a)(x-b)}\right]=\dfrac{x^2}{(x-a)(x-b)}$$
$$1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}=\cdots=\dfrac{x^3}{(x-a)(x-b)(x-c)}$$
A: We can reduce this to all be over the denominator $(x-a)(x-b)(x-c)$:
\begin{align}y&=1 + \frac{a}{x-a} + \frac{bx}{(x-a)(x-b)} + \frac{cx^2}{(x-a)(x-b)(x-c)}\\\\
&=\frac{(x-a)(x-b)(x-c)}{(x-a)(x-b)(x-c)}+\frac{a(x-b)(x-c)}{(x-a)(x-b)(x-c)}\\
&\qquad\qquad+\frac{bx(x-c)}{(x-a)(x-b)(x-c)}+\frac{cx^2}{(x-a)(x-b)(x-c)}\\\\
&=\frac{(x-a)(x-b)(x-c)+a(x−b)(x−c)+bx(x−c)+cx^2}{(x-a)(x-b)(x-c)}\\\\
&=\frac{-a b c + a b x + a c x - a x^2 + b c x - b x^2 - c x^2 + x^3 +ab c - ab x - ac x + ax^2 +bx^2-bcx+cx^2}{(x-a)(x-b)(x-c)}\\\\
&=\frac{x^3+(-a-b-c+a+b+c)x^2+(ab+ac+bc-ab-ac-bc)x+(-abc+abc)}{(x-a)(x-b)(x-c)}\\
&=\frac{x^3}{(x-a)(x-b)(x-c)}\end{align}
Now, we know that $(x-a)(x-b)(x-c)=x^3−3x^2+2x+4$, and so we have $$y=\frac{x^3}{x^3-3x^2+2x+4}$$
Therefore, when $x=2$, \begin{align}y&=\frac{2^3}{2^3-3\times 2^2+2\times 2+4}\\
&=\frac{8}{8-12+4+4}\\
&=\frac{8}{4}\\
&=2\end{align}
A: $(x-a)(x-b)(x-c)+a(x-b)(x-c)+bx(x-c)+cx^2=x^3,$
which says that
$$y=\frac{x^3}{(x-a)(x-b)(x-c)}.$$
Thus, $$y(2)=\frac{8}{(2-a)(2-b)(2-c)}=\frac{8}{8-4\cdot3+2\cdot2-(-4)}=2.$$
A: At $x=2$ we have 
$$y=\frac{4 c}{(2-a) (2-b) (2-c)}+\frac{2 b}{(2-a) (2-b)}+\frac{a}{2-a}+1=-\frac{8}{(a-2) (b-2) (c-2)}$$
Which is
$$y=\frac{8}{-a b c-4 (a+b+c)+2 (a b+a c+b c)+8}$$
In the given  equation $x^3 -3x^2 + 2x +4 = 0$ we know the roots $a,b,c$ so we can write $(x-a)(x-b)(x-c)=0$ that is $$x^3-x^2 (a+b+c)+x (a b+a c+b c)-a b c=0$$
Thus, comparing with the equation's coefficient, we get
$$a+b+c=3;\;a b+a c+b c=2;\;abc=-4$$
Substitute in the expression for $y$ and get
$$y=\frac{8}{4-4 \cdot 3+2 \cdot 2+8}=2$$
