[tex]\text{x}^3-3\text{x}+2=0\\ \\ $Stim ca, solutiile \text{x}_1, \text{x}_2$ si $ \text{x}_3 $ verifica ecuatia, adica, inlocuind cu \\ \text{x}_1, \text{x}_2$ sau $ $cu $ \text{x}_1$ in ecuatie, acea expresie va fi egala cu 0, facem \\ sistem: \\ \\ \left\{ \begin{array}{c} \text{x}_1^3-3\text{x}_1+2 = 0 \\ \text{x}_2^3-3\text{x}_2+2 = 0\\ \text{x}_3^3-3\text{x}_3+2 = 0 \end{array} \right |_{\Big{\text{adunam (+)}}}\Rightarrow\\ \\[/tex]
[tex] \Rightarrow \text{x}_1^3+ \text{x}_2^3+ \text{x}_3^3-3 \text{x}_1-3 \text{x}_2-3 \text{x}_3+2+2+2 = 0 \Rightarrow \\ \\ \Rightarrow \text{x}_1^3+ \text{x}_2^3+ \text{x}_3^3-3( \text{x}_1+ \text{x}_2+ \text{x}_3)+6 = 0 \Rightarrow \\ \\ \Rightarrow \text{x}_1^3+ \text{x}_2^3+ \text{x}_3^3 = 3( \text{x}_1+ \text{x}_2+ \text{x}_3)-6 \\ \\ $Din relatiile lui Viete: \text{x}_1+ \text{x}_2+ \text{x}_3= -\dfrac{b}{a} = -\dfrac{0}{1} = 0 [/tex]
[tex]\Rightarrow \text{x}_1^3+ \text{x}_2^3+ \text{x}_3^3 =3\cdot 0-6 \Rightarrow \boxed{\text{x}_1^3+ \text{x}_2^3+ \text{x}_3^3 = -6}[/tex]
