a) Restul impartirii la binomul x-a este egal cu f(a)deci avem f(1)=8
1+1-m=8, m=-6
[tex]x_{1}+x_{2}+x_{3}=0\\
x_{1}x_{2}+x_{2}x_{3}+x_{1}x_{3}=1\\
x_{1}x_{2}x_{3}=m[/tex]
b)[tex] x_{1}^2+ x_{2}^2+x_{3}^2=(x_{1}+x_{2}+x_{3})^2-2(x_{1}x_{2}+x_{2}x_{3}+x_{1}x_{3})=\\
0-2\cdot 1=-2[/tex]∈Z
c) In conditiile date fac substitutia x→[tex] \frac{1}{y} [/tex]
[tex] \frac{1}{y^3} + \frac{1}{y} -2=0\\
-2y^3+y^2+1=0
f=-2x^3+x^2+1\\
a=-2, b=1, c=0,d=1
[/tex]
