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| vieta,s formula https://forumdematematica.org/viewtopic.php?f=22&t=143 |
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| Autor: | kinu [ 18 jan 2012, 13:53 ] |
| Título da Pergunta: | vieta,s formula |
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 |
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| Autor: | João P. Ferreira [ 23 jan 2012, 00:15 ] |
| Título da Pergunta: | Re: vieta,s formula |
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 |
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| Autor: | kinu [ 26 jan 2012, 08:49 ] |
| Título da Pergunta: | Re: vieta,s formula |
Thanks Professor I will try myself |
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