vieta,s problem

[jagdish]

If $\alpha,\beta,\gamma$ be the roots of the equation $x^3-3x+1=0$

The find value of $(\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)=$

[sourav das]

Denote the roots as $a,b,c$
Now, $a+b+c=0....(i)$; $ab+bc+ca=-3.....(ii)$; $abc=-1....(iii)$

So using (iii) we'll get $(a-b)(b-c)(c-a)=\sum_{cyc} \frac{a}{c} - \sum_{cyc} \frac{a}{b}$
Set $\sum_{cyc} \frac{a}{c}=A$; $\sum_{cyc} \frac{a}{b}=B$ Now we need to find $A-B$
Using (i) $A+B=-3$
Note that, using (i) $9=(ab+bc+ca)^2=ab^2+bc^2+ca^2$ ...(iv)
Same way $(ab)^3+(bc)^3+(ca)^3=(ab+bc+ca)((ab)^2+(bc)^2+(ca)^2)+3a^2b^2c^2=-24$....(v)
As $a+b+c=0$, $a^3+b^3+c^3=3abc=-3$....(vi)
Now using (iii),(v),(vi) $AB=3+\sum_{cyc}\frac{a^2}{bc} + \sum_{cyc}\frac{bc}{a^2}=3+3-24=-18$

So, $(A-B)^2=(A+B)^2-4AB=9+72=81$ and so, $A-B= \pm 9$

[jagdish]

Thanks Sourav