Question

puteti sa mi explicati rezolvarea asta ?
unde x₁ x₂ si x₃ sunt solutiile ecuatiei x³-3x +2=0
sa se arate ca suma radacinilor este -6

Answer 1

Ecuatia se descompune in:
(x-1)(x²+x-2)=0
x-1=0 rezulta x1=1
x²+x-2=0
∆=1+8 rezulta √∆=3
x2=(-1+3)/2=1, x3=(-1-3)/2=-2
1³+1³+(-2)³=-6 q.e.d.

Answer 2

$\text{x}^3-3\text{x}+2=0\\ \\ Stim ca, solutiile \text{x}_1, \text{x}_2 si \text{x}_3 verifica ecuatia, adica, inlocuind cu \\ \text{x}_1, \text{x}_2 sau cu \text{x}_1 in ecuatie, acea expresie va fi egala cu 0, facem \\ sistem: \\ \\ \left\{ \begin{array}{c} \text{x}_1^3-3\text{x}_1+2 = 0 \\ \text{x}_2^3-3\text{x}_2+2 = 0\\ \text{x}_3^3-3\text{x}_3+2 = 0 \end{array} \right |_{\Big{\text{adunam (+)}}}\Rightarrow$

$\Rightarrow \text{x}_1^3+ \text{x}_2^3+ \text{x}_3^3-3 \text{x}_1-3 \text{x}_2-3 \text{x}_3+2+2+2 = 0 \Rightarrow \\ \\ \Rightarrow \text{x}_1^3+ \text{x}_2^3+ \text{x}_3^3-3( \text{x}_1+ \text{x}_2+ \text{x}_3)+6 = 0 \Rightarrow \\ \\ \Rightarrow \text{x}_1^3+ \text{x}_2^3+ \text{x}_3^3 = 3( \text{x}_1+ \text{x}_2+ \text{x}_3)-6 \\ \\ Din relatiile lui Viete: \text{x}_1+ \text{x}_2+ \text{x}_3= -\dfrac{b}{a} = -\dfrac{0}{1} = 0$

$\Rightarrow \text{x}_1^3+ \text{x}_2^3+ \text{x}_3^3 =3\cdot 0-6 \Rightarrow \boxed{\text{x}_1^3+ \text{x}_2^3+ \text{x}_3^3 = -6}$