\(4x^2-24x+36=\left(x-3\right)^3\)
\(\Leftrightarrow4x^2-24x+36=x^3-9x^2+27x-27\)
\(\Leftrightarrow-x^3+13x^2-51x+63=0\)
\(\Leftrightarrow\left(-x^3+10x^2-21x\right)+\left(3x^2-30x+63\right)=0\)
\(\Leftrightarrow-x\left(x^2-10x+21\right)+3\left(x^2-10x+21\right)=0\)
\(\Leftrightarrow\left(x^2-10x+21\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left(x^2-3x-7x+21\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left[x\left(x-3\right)-7\left(x-3\right)\right]\left(3-x\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-7\right)\left(3-x\right)=0\)
\(\Leftrightarrow\left(3-x\right)^2\left(7-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\left(3-x\right)^2=0\\7-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}}\)
Vậy...
\(4x^2-24x+36=\left(x-3\right)^3\)\(\Leftrightarrow4\left(x^2-6x+9\right)=\left(x-3\right)^3\)
\(\Leftrightarrow4\left(x-3\right)^2=\left(x-3\right)^3\)\(\Leftrightarrow4\left(x-3\right)^2-\left(x-3\right)^3=0\)
\(\Leftrightarrow\left(x-3\right)^2\left[4-\left(x-3\right)\right]=0\)\(\Leftrightarrow\left(x-3\right)^2\left(4-x+3\right)=0\)
\(\Leftrightarrow\left(x-3\right)^2\left(7-x\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}\left(x-3\right)^2=0\\7-x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x-3=0\\x=7\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=7\end{cases}}\)
Vậy \(x=3\)hoặc \(x=7\)
