[Raiz Cúbica] Diferença de Raízes Cúbicas

por **CJunior** » Sex Fev 28, 2014 21:31

( IME 1991) Mostre que $\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}$ é um número racional.

por **young_jedi** » Sáb Mar 01, 2014 13:40

$x=\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}$

$x^3=\left(\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}\right)^3$

$x^3=3+\sqrt[2]{9+\frac{125}{27}}-3\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}^2.\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}+$ $3\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}^2+3-\sqrt[2]{9+\frac{125}{27}}$

$x^3=6-3\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}^2.\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}+ 3\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}^2$

$x^3=6-3\sqrt[3]{\left(3+\sqrt[2]{9+\frac{125}{27}}\right)^2.\left(-3+\sqrt[2]{9+\frac{125}{27}}\right)}+$ $3\sqrt[3]{\left(3+\sqrt[2]{9+\frac{125}{27}}\right).\left(-3+\sqrt[2]{9+\frac{125}{27}}\right)^2}$

$x^3=6-3\sqrt[3]{\left(3+\sqrt[2]{9+\frac{125}{27}}\right).\left(-3^2+\sqrt[2]{9+\frac{125}{27}}^2\right)}$ $+3\sqrt[3]{\left(-3^2+\sqrt[2]{9+\frac{125}{27}}^2\right).\left(-3+\sqrt[2]{9+\frac{125}{27}}\right)}$

$x^3=6-3\sqrt[3]{\left(3+\sqrt[2]{9+\frac{125}{27}}\right).\left(-9+9+\frac{125}{27}\right)}$ $+3\sqrt[3]{\left(-9+9+\frac{125}{27}\right).\left(-3+\sqrt[2]{9+\frac{125}{27}}\right)}$

$x^3=6-3\sqrt[3]{\left(3+\sqrt[2]{9+\frac{125}{27}}\right).\frac{125}{27}}$ $+3\sqrt[3]{\frac{125}{27}.\left(-3+\sqrt[2]{9+\frac{125}{27}}\right)}$

$x^3=6-3.\frac{5}{3}\sqrt[3]{\left(3+\sqrt[2]{9+\frac{125}{27}}\right)}$ $+3.\frac{5}{3}\sqrt[3]{\left(-3+\sqrt[2]{9+\frac{125}{27}}\right)}$

$x^3=6-5\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}$ $+5\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}$

$x^3=6-5\left(\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}\right)$

x^3=6-5x

é facil ver que a raiz real dessa equação é 1

portanto

$x=\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}=1$

por **Man Utd** » Ter Mar 04, 2014 15:27

CJunior escreveu:( IME 1991) Mostre que $\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt[2]{9+\frac{125}{27}}}$ é um número racional.

$x=\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}-\sqrt[3]{-\left(3-\sqrt[2]{9+\frac{125}{27}}\right)}$

$x=\sqrt[3]{3+\sqrt[2]{9+\frac{125}{27}}}+\sqrt[3]{3-\sqrt[2]{9+\frac{125}{27}}}$

Perceba que agora está no "jeito" da fórmula de cardano : $x=\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^{2}}{4}+\frac{p^{3}}{27}}}$
que serve para resolver equações cúbicas reduzidas do tipo: $x^{3}+px+q=0$ .Enfim comparando-a com a fórmula obtemos : $q=-6 \;\; \wedge \;\; p=5$ ,segue que a equação é :

x^3+5x-6=0

que já sabemos que a raiz real é $\boxed{\boxed{1}}$