[tex]\displaystyle\\\
\text{Rezolvam ecuatia de gradul 3.}\\\\
x^3-3x+2=0\\\\
\text{Printre divizorii termenului liber cautam o solute a ec.}\\\\
\text{Observam ca 1 verifica ecuatia:}~1^3-3\cdot1+2=0\\\\
x^3-x^2+x^2-x-2x+2=0\\\\
x^2(x-1)+x(x-1)-2(x-1)=0\\\\
(x-1)(x^2+x-2)=0 \\\\
x_1=\boxed{1}\\\\
x_{23}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{-1\pm\sqrt{1+4\cdot2}}{2}=\frac{-1\pm\sqrt{9}}{2}=\frac{-1\pm3}{2}\\\\
x_2=\frac{-1+3}{2}=\frac{2}{2}=\boxed{1}\\\\
x_3=\frac{-1-3}{2}=\frac{-4}{2}=\boxed{-2}[/tex]
[tex]\displaystyle\\
\text{Avem solutiile: } \\
x_1 = 1\\
x_2 = 1\\
x_3 = -2 \\\\
a)\\
x_1+x_2+x_3 = 1+1-2 = \boxed{0}\\\\
b)\\
x_1^3 +x_2^3 + x_3^3 = 1^3 + 1^3 + (-2)^3 = 1+1-8 = \boxed{-6} \\\\
c)\\
d = \left[\begin{array}{ccc}x_1&x_2&x_3\\x_2&x_3&x_1\\ x_3&x_1&x_2\end{array}\right] = \left[\begin{array}{ccc}1&1&-2\\1&-2&1\\-2&1&1\end{array}\right]=\\\\\\
=-2 -2-2 +8 -1 -1 = \boxed{0}
[/tex]
