Solve for x $\left\lfloor x \right\rfloor^3+2x^2=x^3+2\left\lfloor x\right\rfloor^2$ Solve for x 
$$\left\lfloor x \right\rfloor^3+2x^2=x^3+2\left\lfloor x\right\rfloor^2$$
where $\left\lfloor t \right\rfloor$ denotes the largest integer not exceeding t
$X \in \mathbb{Z}$ is a solution. Is there other root? Thanks.

Write the equation as $x^3 - 2x^2 = \lfloor x \rfloor^3 - 2\lfloor x \rfloor^2$. Clearly this has solutions in $\mathbb{Z}$.
On most intervals of the form $[n, n+1]$ with $n \in \mathbb{Z}$, the function $x^3 - 2x^2$ is monotone. You will only get interesting values on the intervals where it isn't.
To find those, take the derivativve $3x^2 - 4x$ and set it to zero - you get $0$ and $\frac{4}{3}$ as critical points. So the only possible zero outside of $\mathbb{Z}$ will occur where $\lfloor x \rfloor = 1$.
Solve the equation now: $x^3 - 2x^2 = -1$ has three real solutions, one of which lies in the interval $(1, 2)$ and therefore satisfies $\lfloor x \rfloor = 1$.
This solution is $\frac{1}{2}(1 + \sqrt{5})$.
