vieta,s formula

[kinu]

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

The value of \((\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)=\) is

[João P. Ferreira]

Hi

First of all i recommend you to read this

http://en.wikipedia.org/wiki/Vieta's_formulas

On this polynomial we know that

\(x^3-3x+1=0\)

\(\begin{cases}
a_3=1\\
a_2=0\\
a_1=-3\\
a_0=1\\
\end{cases}\)

Using Vieta's formula we know that

\(\begin{cases}
\alpha+\beta+\gamma=-\frac{a_2}{a_3}\\
\alpha\beta+\alpha\gamma+\beta\gamma=\frac{a_1}{a_3}\\
\alpha\beta\gamma=(-1)^3 . \frac{a_0}{a_3}\\
\end{cases}\)

We know then:

\(\begin{cases}
\alpha+\beta+\gamma=0\\
\alpha\beta+\alpha\gamma+\beta\gamma=-3\\
\alpha\beta\gamma=-1\\
\end{cases}\)

you just need to try to find now the expression you want

[kinu]

Thanks Professor I will try myself