Question

B = \(x^3+\dfrac{1}{x^3}\), biết rằng \(x^2+\dfrac{1}{x^2}=7\)

Answer 1

\(x^2+\dfrac{1}{x^2}=7\\ \\ \\ \\ \Rightarrow x^2+2x\cdot\dfrac{1}{x}+\dfrac{1}{x^2}=7+2x\cdot\dfrac{1}{x}=9\\ \\ \\ \\ \Rightarrow\left(x+\dfrac{1}{x}\right)^2=9\\ \\ \\ \\ \Rightarrow x+\dfrac{1}{x}=\pm3\)

Trường hợp 1: \(x+\dfrac{1}{x}=3\) ta có:

\(x^3+\dfrac{1}{x^3}=\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)-\left(x+\dfrac{1}{x}\right)=7\cdot3-3=18\)

Trường hợp 2: \(x+\dfrac{1}{x}=-3\) ta có:

\(x^3+\dfrac{1}{x^3}=\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)-\left(x+\dfrac{1}{x}\right)=7\cdot\left(-3\right)-\left(-3\right)=-18\)