[tex]f(x) = x^3+ax+b\\ \\ f'(x) = 3x^2+a\\ f'(x) = 0 \Rightarrow 3x^2+a = 0 \Rightarrow 3x^2 = -a \Rightarrow x^2 = -\dfrac{a}{3} \Rightarrow \\ \Rightarrow x_1 = \sqrt{-\dfrac{a}{3}},\quad x_2 = -\sqrt{-\dfrac{a}{3}}\\ \\ \text{Nu ne intereseaza care e abscisa minimului sau maximului,}\\\text{stim doar ca una din ele e cea a minimului iar cealalta a maximului.}\\ \\ m\cdot M = f\Big(\sqrt{-\dfrac{a} {3}}\Big)\cdot f\Big(-\sqrt{-\dfrac{a}{3}}\Big) =[/tex]
[tex]= \Big(\sqrt{-\dfrac{a}{3}}^3+a\cdot \sqrt{-\dfrac{a}{3}}+b\Big)\Big(-\sqrt{-\dfrac{a}{3}}^3-a\cdot \sqrt{-\dfrac{a}{3}}+b\Big) = \\ \\ = b^2 - \Big(\sqrt{-\dfrac{a}{3}}^3+a\cdot \sqrt{-\dfrac{a}{3}}\Big)^2 = \\ \\ = b^2-\Big[\Big(-\dfrac{a}{3}\Big)^3+2a\cdot \Big(-\dfrac{a}{3}\Big)^2+a^2\cdot \Big(-\dfrac{a}{3}\Big)\Big] = \\ \\ = b^2+\dfrac{a^3}{27}-\dfrac{2a^3}{9}+\dfrac{a^3}{3} = \\ \\ = b^2+\dfrac{a^3-6a^3+9a^3}{27} = \\ \\ = \boxed{b^2+\dfrac{4}{27}a^3}[/tex]
