Question

Se x+1/x=3, calcule o valor de x ao cubo +1/x ao cubo.

Answer 1

$\boxed{\boxed{\mathsf{(a+b)^{3}=a^{3}+3a^{2}b+3ab^{2}+b^{3}}}}$

Como $x+\frac{1}{x}=3~~\Longrightarrow~~(x+\frac{1}{x})^{3}=3^{3}=27$

Agora, vamos expandir $(x+\frac{1}{x})^{3}$:

$27=(x+\frac{1}{x})^{3}=x^{3}+3x^{2}(\frac{1}{x})+3x(\frac{1}{x})^{2}+(\frac{1}{x})^{3}$

Rearrumando a equação, temos

$x^{3}+3x^{2}\cdot\frac{1}{x}+3x\cdot\frac{1}{x^{2}}+(\frac{1}{x})^{3}=27\\\\x^{3}+3x+3\cdot\frac{1}{x}+(\frac{1}{x})^{3}=27$

Colocando 3 em evidência nos membros centrais:

$x^{3}+3\cdot(x+\frac{1}{x})+(\frac{1}{x})^{3}=27$

Substituindo $x+\frac{1}{x}=3$:

$x^{3}+3\cdot3+(\frac{1}{x})^{3}=27\\\\x^{3}+9+(\frac{1}{x})^{3}=27\\\\x^{3}+(\frac{1}{x})^{3}=27-9\\\\x^{3}+(\frac{1}{x})^{3}=18$

Então:

$\displaystyle\boxed{\boxed{x^{3}+\bigg(\frac{1}{x}\bigg)^{3}=18}}$