`x^3+1/x^3=6(x+1/x)(x ne 0)`
`<=>(x+1/x)(x^2-1+1/x^2)=6(x+1/x)`
`<=>(x+1/x)(x^2-1+1/x^2-6)=0`
`<=>((x^2+1)/x)(x^2+1/x^2-7)=0`
`(x^2+1)/x ne 0(AA x)`
`=>x^2+1/x^2-7=0`
`=>x^2+2+1/x^2-9=0`
`<=>(x+1/x)^2-3=0`
`<=>(x+1/x+3)(x+1/x-3)=0`
`+)x+1/x+3=0`
`<=>(x^2+3x+1)/x=0`
`<=>x^2+3x+1=0`
`<=>x^2+3x+9/4=5/4`
`<=>(x+3/2)^2=5/4`
`<=>x=(+-\sqrt{5}-3)/2`
`+)x+1/x-3=0`
`<=>(x^2-3x+1)/x=0`
`<=>x^2-3x+1=0`
`<=>x^2-3x+9/4=5/4`
`<=>(x-3/2)^2=5/4`
`<=>x=(+-\sqrt{5}+3)/2`
Vậy `S={(\sqrt{5}-3)/2,(-\sqrt{5}-3)/2,(\sqrt{5}+3)/2,(-\sqrt{5}+3)/2}`
ĐK: \(x\ne0\)
\(PT\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(x^2-1+\dfrac{1}{x^2}\right)=6\left(x+\dfrac{1}{x}\right)\)
\(\Leftrightarrow\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{x}=0\\x^2+\dfrac{1}{x^2}-7=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(loai\right)\\x^4-7x^2+1=0\left(1\right)\end{matrix}\right.\)
Giải (1): \(\Leftrightarrow x^2=\dfrac{7\pm3\sqrt{5}}{2}\) \(\Rightarrow x=\pm\sqrt{\dfrac{7\pm3\sqrt{5}}{2}}\)