Question

Nu-mi da deloc. Îmi dă m patrat- 6

Answer 1

$f = x^3-mx+1 \\ \\ x^3-mx+2 = 0 \Rightarrow x^3 = mx-2 \Rightarrow \\ \\\Rightarrow x_k^3 = mx_k-2,\quad k =\overline{1,3} \\ \\\\ \Rightarrow \sum\limits_{k=1}^3x_k^3 = \sum\limits_{k=1}^3 (mx_k-2) \Rightarrow \\\ \\ \Rightarrow x_1^3+x_2^3+x_3^3 = m(x_1+x_2+x_3)-2\cdot 3 = m\cdot \Big(-\dfrac{0}{1}\Big)-6 = \\ \\ = 0-6 = \boxed{-6}$