<s>[INST] 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!
 [/INST] 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 $3n+3=6\to n=1$.
$m=1*2+2*3+3*1=11$
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