Series question involving a cubic polynomial The question asks: Consider the polynomial
$\displaystyle{\,\mathrm{f}\left(X\right) = X^{3} -6X^{2} + mX - 6}$, where $m$ is a real parameter. 
a. Show that:
$\displaystyle{{1 \over x_{1}x_{2}} + {1 \over x_{1}x_{3}} + {1 \over x_{2}x_{3}} = 1}$ where $\displaystyle{x1,x2,x3}$ are the roots of the polynomial $\,\mathrm{f}$ . 
b. Determine the parameter $m$ such that the roots of polynomial $f$ are three consecutive integer numbers  
For part a., I have attempted to solve it as follows:  
$\displaystyle{\,\mathrm{f} = X\left(X^{2} - 6X + m\right) - 6 = 0}$
$(X-6)(X^2-6X+m)=0$
$x1=6$  
$X^2-6X+m=0$
Solving quadratically, $x2=3+\sqrt{36-4m}$,
$x3=3-\sqrt{36-4m}$  
Then I substituted these roots into the equation to be proved. However, I get into really messy workings, ending up with the wrong answer.  
Could someone please guide me on solving both part (a) and (b)? It would be highly appreciated!
 A: 
$f=X(X^2-6X+m)-6=0$
  $(X-6)(X^2-6X+m)=0$ 

This is incorrect since $X(X^2-6X+m)-6\not=(X-6)(X^2-6X+m)$.

Using Vieta's formulas should help :
$$x_1+x_2+x_3=6,\quad x_1x_2+x_2x_3+x_3x_1=m,\quad x_1x_2x_3=6$$
For a : 
$$\frac{1}{x_1x_2}+\frac{1}{x_1x_3}+\frac{1}{x_2x_3}=\frac{x_1+x_2+x_3}{x_1x_2x_3}$$
For b : 
We may suppose that $x_1=k,x_2=k+1,x_3=k+2$, so
$$6=k+(k+1)+(k+2)\implies k=1$$
from which we can get $m$.
A: If $x_1,x_2,x_3$ are the roots of the polynomial $ p(x)=x^3-6x^2+mx-6$, then $z_1=\frac{1}{x_1},z_2=\frac{1}{x_2},z_3=\frac{1}{x_3}$ are the roots of the "reciprocal polynomial"
$$ q(z) = -\frac{z^3}{6}\cdot p\!\left(\frac{1}{z}\right) =z^3-\frac{m}{6}z^2+z-\frac{1}{6}\tag{1}$$
and by Vieta's formulas
$$ z_1 z_2+z_1 z_3+z_2 z_3 = e_2(z_1,z_2,z_3) = \frac{1}{1}=\color{red}{1}\tag{2}$$
no matter what the value of $m$ is.
A: Let a,b,c the roots. Note that $\frac{1}{ab}+ \frac{1}{ac}+ \frac{1}{bc}= \frac{a+b+c}{abc}=*$.
Now, $f=(x-a)(x-b)(x-c)=x^3-(a+b+c)x^2+(ab+bc+ca)x-abc$.
And then $*=6/6=1$.
For b, note that $m=ab+bc+ca$
If (suposse) $n, n+1, n+2$ are the roots, their sum is 6 by a), and then $3n+3=6\to n=1$.
$m=1*2+2*3+3*1=11$
