$f(x)$ is a polynomial of degree three which leaves remainder $1$ when divided by $(x−1)^2$ and leaves remainder $–1$ when divided by $(x+1)^2$.
If $f(x)=0$ has roots $\alpha,\beta,\gamma$, then $$(\alpha\beta+\beta\gamma+\gamma\alpha)=\text?$$
The answer to the question is $-3$. How do I approach this problem?
Note that $P'(x)$ has roots $1,-1$. Hence $P'(x) = a(x-1)(x+1) =a (x^2-1)$
and hence $P(x) = a \left(\dfrac{x^3}{3} -x\right)+b$. Using $P(1)=1, P(-1)=-1$, we get $b=0, a=-\dfrac{3}{2}$
Hence $P(x) =\dfrac{x^3-3x}{2}$. It is easy to see that $\displaystyle \sum_{cyc} \alpha \beta= -3$
Hint. Your first assumption says that $(f-1)(1)=(f-1)'(1)=0$ since $f-1$ is divisible by $(X-1)^2,$ and your second assumption says that $ (f+1)(-1)=(f+1)'(-1)=0$ since $f+1$ is divisible by $(X+1)$^2.
You then get a linear system that you can solve to finally find $f$. Then apply roots-coefficients relations.
Try to see how to write explicit polynomial equations, i.e., $$ f(x)=(ax+b)(x-1)^2+1=(cx+d)(x+1)^2-1 $$ Comparison of the $4$ coefficients of the two cubic polynomials gives you $4$ linear equations in $a,b,c,d$ with a unique solution. The resulting cubic polynomial $f$ is very easy and has only two terms.
Hint
Let $f(x)=1+(x-1)^2(ax+b)=-1+(x+1)^2(cx+d)$
$ax^3+x^2(b-2a)+x(a-2b)+b-1=cx^3+x^2(d+2c)+x(c+2d)+d+1$
Now compare the coefficients of the different exponent of $x$
$\implies c=a$
$b-1=d+1\iff d=b-2$
$b-2a=d+2c\iff b=d+2(c+a)=d+4a$
$a-2b=c+2d\implies b=-d=2-b$
Hope you can take it from here?
$f\!+\!1\bmod{(x\!+\!1)^2(x\!-\!1)^2} = (x\!+\!1)^2\left[\dfrac{\smash{\overbrace{f\!+\!1}^{\textstyle{\color{#c00}1\!+\!1}}}}{(x\!+\!1)^2}\,\bmod{(x\!-\!1)^2}\right]$ $=\, (x\!+\!1)^2\left[\dfrac{\color{#0a0}{2\!-\!x}}{2}\right]$
by, with $t=x\!-\!1,\,$ we have: $ \bmod t^2\!:\,\ \dfrac{4}{(x\!+\!1)^2} = \dfrac{4}{(2\!+\!t)^2} =\dfrac{1}{1\!+\!t} = 1\!-\!t = \color{#0a0}{2\!-\!x}$
