\(\left(x^3-27\right):\left(x-3\right)=\left(x-1\right)^2.\left(x+3\right)\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x^2+3x+9\right)}{x-3}=\left(x^2-2x+1\right)\left(x+3\right)\) ĐKXĐ:\(x\ne3\)
\(\Leftrightarrow x^2+3x+9=x^3+3x^2-2x^2-6x+x+3\)
\(\Leftrightarrow x^2+3x+9=x^3+x^2-5x+3\)
\(\Leftrightarrow-x^3+x^2-x^2+5x=-9+3\)
\(\Leftrightarrow-x^3+5x+6=0\)
tới đây me bí rồi