Question

sendo > 0 e x2+ 1/x2 = 5 , determine x3 + 1/x3

Answer 1

Resposta: 4√7

Explicação passo-a-passo:

Para x > 0:

$\tt x^2+\dfrac{1}{x^2}=5\\\\\\ \tt x^2+2+\dfrac{1}{x^2}=5+2\\\\\\ \tt x^2+2\cdot x\cdot \dfrac{1}{x}+\dfrac{1}{x^2}=7\\\\\\ \tt x^2+2\cdot x\cdot \dfrac{1}{x}+\left(\dfrac{1}{x}\right)^{\!\!2}=7\\\\\\ \tt \left(x+\dfrac{1}{x}\right)^{\!\!2}=7\\\\\\ \tt x+\dfrac{1}{x}=\sqrt{7}$

Portanto:

$\tt \left(x+\dfrac{1}{x}\right)^{\!\!3}=x^3+3\cdot x^2\cdot \dfrac{1}{x}+3\cdot x\cdot \left(\dfrac{1}{x}\right)^{\!\!2}+\left(\dfrac{1}{x}\right)^{\!\!3}\\\\\\ \left(x+\dfrac{1}{x}\right)^{\!\!3}=x^3+3\:\!x+\dfrac{3}{x}+\dfrac{1}{x^3}\\\\\\ \tt \left(x+\dfrac{1}{x}\right)^{\!\!3}=x^3+\dfrac{1}{x^3}+3\cdot \underbrace{\tt \!\left(x+\dfrac{1}{x}\right)}_{\tt \sqrt{7}}\\\\\\ \tt \underbrace{\tt \left(x+\dfrac{1}{x}\right)^{\!\!3}}_{\sqrt{7^3}}=x^3+\dfrac{1}{x^3}+3\sqrt{7}$

$\tt x^3+\dfrac{1}{x^3}=\sqrt{7^3}-3\sqrt{7}\\\\\\ \tt x^3+\dfrac{1}{x^3}=7\sqrt{7}-3\sqrt{7}\\\\\\ \tt \boxed{\tt x^3+\dfrac{1}{x^3}=4\sqrt{7}}$