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!

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 $3n+3=6\to n=1$.
$m=1*2+2*3+3*1=11$
