## vieta,s problem

For discussing Olympiad Level Algebra (and Inequality) problems
jagdish
Posts: 38
Joined: Wed Jan 19, 2011 2:21 pm

### vieta,s problem

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

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

sourav das
Posts: 461
Joined: Wed Dec 15, 2010 10:05 am
Location: Dhaka
Contact:

### Re: vieta,s problem

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, $A-B= \pm 9$
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

jagdish
Posts: 38
Joined: Wed Jan 19, 2011 2:21 pm