Resolução
[tex3]\bullet x^{3}-x-1=0[/tex3]
[tex3]\bullet [/tex3]
Relações de Girard:
[tex3]\alpha +\beta +\theta =0[/tex3]
[tex3]\alpha \beta +\alpha \theta +\beta \theta =-1[/tex3]
[tex3]\alpha \beta \theta =1[/tex3]
[tex3]\bullet M=\frac{1+\alpha }{1-\alpha }+\frac{1+\beta }{1-\beta }+\frac{1+\theta }{1-\theta }[/tex3]
[tex3]M=(\frac{1+\alpha }{1-\alpha }+1)+(\frac{1+\beta }{1-\beta }+1)+(\frac{1+\theta }{1-\theta }+1)-3[/tex3]
[tex3]M=\frac{1+\alpha +1-\alpha }{1-\alpha }+\frac{1+\beta +1-\beta }{1-\beta }+\frac{1+\theta +1-\theta }{1-\theta }-3[/tex3]
[tex3]M=\frac{2}{1-\alpha }+\frac{2}{1-\beta }+\frac{2}{1-\theta }-3[/tex3]
[tex3]M=2(\frac{1}{1-\alpha }+\frac{1}{1-\beta }+\frac{1}{1-\theta })-3[/tex3]
[tex3]M=2.\frac{(1-\alpha )(1-\beta )+(1-\alpha )(1-\theta )+(1-\beta )(1-\theta )}{(1-\alpha )(1-\beta )(1-\theta )}-3[/tex3]
[tex3]M=2.\frac{3-2(\alpha +\beta +\theta )+(\alpha \beta +\alpha \theta +\beta \theta )}{1-(\alpha +\beta +\theta )+(\alpha \beta +\alpha \theta +\beta \theta )-\alpha \beta \theta }-3[/tex3]
Substituindo os valores:
[tex3]M=2.\frac{3-2.0-1}{1-0-1-1}-3[/tex3]
[tex3]M=2.(-2)-3[/tex3]
[tex3]\therefore \boxed{M=-7}[/tex3]