Ta thấy:
\(\left(2x-1\right)+\left(3-x\right)+\left(-2-x\right)\\ =2x-1+3-x-2-x\\ =\left(2x-x-x\right)+\left(3-1-2\right)\\ =0\)
Ta có:
\(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)
\(\Rightarrow\left(2x-1\right)^3+\left(3-x\right)^3+\left(-2-x\right)^3=0\\ \Leftrightarrow3\left(2x-1\right)\left(3-x\right)\left(-2-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x-1=0\\3-x=0\\-2-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0,5\\x=3\\x=-2\end{matrix}\right.\)
Vậy \(x=0,5;x=3\) hoặc \(x=-2\)
Áp dụng x3+y3+z3=3xyz, ta có:
(2x-1)3+(3-x)3+(-2-x)3= 3(2x-1)(3-x)(-2-x)=0
=>\(\left\{{}\begin{matrix}2x-1=0\\3-x=0\\-2-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{1}{2}\\x=3\\x=-2\end{matrix}\right.\)
Vậy nghiệm của pt là x \(\in\left\{-2;\dfrac{1}{2};3\right\}\)