Question

4x-x"'(cubo)=4
estou com dificuldades com essa questão
Mateus 9 ano

Answer 1

$x^3-4x+4=0\\ \\ \text{F\'ormula para: }x^3+px+q=0\\ \\ x=\sqrt[3]{-\dfrac{q}{2}+\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}+\sqrt[3]{-\dfrac{q}{2}-\sqrt{\dfrac{q^2}{4}+\dfrac{p^3}{27}}}\\ \\ \\ x=\sqrt[3]{-2+\sqrt{4-\dfrac{64}{27}}}+\sqrt[3]{-2-\sqrt{4-\dfrac{64}{27}}}\\ \\ \\ \boxed{x=\sqrt[3]{-2+\dfrac{2}{3}\sqrt{\dfrac{11}{3}}}+\sqrt[3]{-2-\dfrac{2}{3}\sqrt{\dfrac{11}{3}}}}$