Suppose $x>30$ is a number satisfying $\lfloor x\rfloor \cdot \lfloor x^2\rfloor = \lfloor x^3\rfloor$. Prove that $\{x\}<\frac{1}{2700}$, where $\{x\}$ is the fractional part of $x$.
My heuristic is that $x$ needs to be "small": i.e. as close to $30$ as possible to get close to the upper bound on $\{x\}$, but I'm not sure how to make this a proof.
Let $\lfloor x \rfloor =y$ and $\{x\}=b$ Then $\lfloor x\rfloor \cdot \lfloor x^2\rfloor = \lfloor x^3\rfloor =y\lfloor y^2+2by+b^2 \rfloor= \lfloor y^3+3y^2b+3yb^2+b^3\rfloor$
One way this can happen is that $b$ is small enough that all the terms including $b$ are less than $1$, which makes both sides $y^3$. This requires $3y^2b \lt 1$, which gives $b \lt \frac 1{2700}$ as required. Now you have to argue that if $2by+b^2 \ge 1$ the right side will be too large.