\(x+\frac{1}{x}=\frac{5}{2}\Leftrightarrow2x^2-5x+2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{1}{2}\end{matrix}\right.\)
- Với \(x=2\Rightarrow x^3-\frac{1}{x^3}=8-\frac{1}{8}=\frac{63}{8}\)
- Với \(x=\frac{1}{2}\Rightarrow x^3-\frac{1}{x^3}=\frac{1}{8}-8=-\frac{63}{8}\)
\(x^3-\frac{1}{x^3}=\left(x-\frac{1}{x}\right)\left(x^2+\frac{1}{x^2}+1\right)=\left(x-\frac{1}{x}\right)\left[\left(x+\frac{1}{x}\right)^2-1\right]\)
\(=\left(x-\frac{1}{x}\right)\left(\frac{25}{4}-1\right)=\frac{21}{4}\left(x-\frac{1}{x}\right)\)
Đặt \(A=x-\frac{1}{x}\Rightarrow A^2=x^2+\frac{1}{x^2}-2=\left(x+\frac{1}{x}\right)^2-4=\frac{25}{4}-4=\frac{9}{4}\)
\(\Rightarrow A=\pm\frac{3}{2}\)
\(\Rightarrow x^3-\frac{1}{x^3}=\frac{21}{4}.\left(\pm\frac{3}{2}\right)=\pm\frac{63}{8}\)
