Solve in real $x,\,\,\lfloor x^3\rfloor-3\lfloor x^2\rfloor+2\lfloor x\rfloor=0$
My progress:
$x$ must be non negative and $x<3$ As $x\ge 3\implies \lfloor x^3\rfloor\ge\lfloor x\rfloor\lfloor x^2\rfloor\ge 3\lfloor x^2\rfloor\implies\lfloor x^3\rfloor-3\lfloor x^2\rfloor+2\lfloor x\rfloor\ge 6$
hence all solutions $\in [0,3)$
For $x\in [0,1),LHS=0-0+0=0$
For $x\in [1,\sqrt[3]2),LHS=1-3+2=0$
For $x\in [2,\sqrt[3]9),LHS=8-12+4=0$