\(x^3=x\)
\(\Rightarrow x^3-x=0\)
\(\Rightarrow x\left(x^2-1\right)=0\)
\(\left\{\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=0\\x=\pm1\end{matrix}\right.\)
Vậy \(x\in\left\{0;1;-1\right\}\)
x3 = x
=> x3 - x = 0
=> x2.x - x = 0
=> x(x2 - 1) = 0
\(\Rightarrow\left\{\begin{matrix}x=0\\x^2-1=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=0\\x^2=1\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
x=0 hoặc x=1 vì
x=x nên x lập phương cũng = x
0*0*0=0
1*1*1=1
