\(\left(x-3\right)^3+\left(2x-3\right)^3=27\left(x-2\right)^3\)
\(\Leftrightarrow\left(x-3+2x-3\right)\left[\left(x-3\right)^2-\left(x-3\right)\left(2x-3\right)+\left(2x-3\right)^2\right]=\left(3x-6\right)^3\)
\(\Leftrightarrow\left(3x-6\right)\left(x^2-6x+9-2x^2+9x-9+4x^2-12x+9\right)=\left(3x-6\right)^3\)
\(\Leftrightarrow\left(3x-6\right)\left(3x^2-9x+9\right)=\left(3x-6\right)\left(9x^2-36x+36\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-6=0\\3x^2-9x+9=9x^2-36x+36\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\6x^2-27x+27=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\3\left(2x-3\right)\left(x-3\right)=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=3\end{matrix}\right.\)
Vậy tập nghiệm của phương trình là : \(S=\left\{2;\frac{3}{2};3\right\}\)
\(\Leftrightarrow\left(x-3\right)^3+\left(2x-3\right)^3=\left(3x-6\right)^3\)
Đặt \(\left\{{}\begin{matrix}x-3=a\\2x-3=b\end{matrix}\right.\)
\(\Rightarrow a^3+b^3=\left(a+b\right)^3\)
\(\Leftrightarrow a^3+b^3=a^3+b^3+3ab\left(a+b\right)\)
\(\Leftrightarrow ab\left(a+b\right)=0\Rightarrow\left[{}\begin{matrix}a=0\\b=0\\a+b=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\2x-3=0\\3x-6=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=3\\x=\frac{3}{2}\\x=2\end{matrix}\right.\)
