Đặt \(\hept{\begin{cases}\left(x+\frac{1}{x}\right)^3=a\\x^3+\frac{1}{x^3}=b\end{cases}}\)
Ta có
\(A=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^6+2+\frac{1}{x^6}\right)}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}=\frac{\left(x+\frac{1}{x}\right)^6-\left(x^3+\frac{1}{x^3}\right)^2}{\left(x+\frac{1}{x}\right)^3+x^3+\frac{1}{x^3}}\)
\(=\frac{a^2-b^2}{a+b}=a-b\)
\(=\left(x+\frac{1}{x}\right)^3-\left(x^3+\frac{1}{x^3}\right)\)
\(=x^3+3\left(x+\frac{1}{x}\right)+\frac{1}{x^3}-\left(x^3+\frac{1}{x^3}\right)=\frac{3x^2+3}{x}\)