Question

Se $k + ( \frac{1}{k} ) = 3$ então $\sqrt{[ k^{3}+ (\frac{1}{k^{3}})] } } = 3 \sqrt{2$

Como é o processo para que esta raiz dê 3V2 ?

Answer 1

Olá, Andrejoau1.

$(k+\frac1 k)^3=(k+\frac1 k)(k+\frac 1 k)^2=(k+\frac1 k)(k^2+2+\frac 1 {k^2})=\\\\k^3+2k+\frac1 k+k+\frac 2 k+\frac 1{k^3}=k^3+3k+\frac 3 k+\frac 1{k^3}=\\\\=k^3+3(k+\frac 1 k)+\frac 1{k^3} \Rightarrow\\\\3^3=k^3+3\cdot3+\frac1{k^3}\Rightarrow k^3+\frac1{k^3}=27-9\\\\\Rightarrow k^3+\frac1{k^3}=18$

Extraindo a raiz quadrada dos dois lados fica:

$\sqrt{k+\frac1 {k^3}}=\sqrt{18}=\sqrt{2\cdot3^2}\Rightarrow\\\\\\\boxed{\sqrt{k+\frac1 {k^3}}=3\sqrt{2}}$