\(\left(x-2\right)^3+\left(3\text{x}-1\right)\left(3\text{x}+1\right)=\left(x+1\right)^3\)
\(\Leftrightarrow\left(x-2\right)^3+\left(3\text{x}-1\right)\left(3\text{x}+1\right)-\left(x+1\right)^3=0\)
\(\Leftrightarrow\left(x^3-6\text{x}^2+12\text{x}-8\right)+\left(9\text{x}^2-1\right)-\left(x^3+3\text{x}^2+3\text{x}+1\right)=0\)
\(\Leftrightarrow x^3-6\text{x}^2+12\text{x}-8+9\text{x}^2-1-x^3-3\text{x}^2-3\text{x}-1=0\)
\(\Leftrightarrow\left(x^3-x^3\right)+\left(-6\text{x}^2+9\text{x}^2-3\text{x}^2\right)+\left(12\text{x}-3\text{x}\right)+\left(-8-1-1\right)=0\)
\(\Leftrightarrow9\text{x}-10=0\)
\(\Leftrightarrow9\text{x}=10\Leftrightarrow x=\frac{10}{9}\)
Vậy x = \(\frac{10}{9}\)
