Solve for real solution of the system below: $\displaystyle \left\lfloor{x^3}\right\rfloor+\left\lfloor{x^2}\right\rfloor+\left\lfloor{x}\right\rfloor=\left\{x\right\}-1$.

Solve for real solution of the system below:

$\displaystyle \left\lfloor{x^3}\right\rfloor+\left\lfloor{x^2}\right\rfloor+\left\lfloor{x}\right\rfloor=\left\{x\right\}-1$.

My solution:

First, observe that the LHS of the equality must yield an integer, this tells us the fractional part of $x$ must be a zero, so this turns the whole equation as:

$\displaystyle x^3+x^2+x=-1\\x^3+x^2+x+1=0\\(x+1)(x^2+1)=0$

This implies $x=-1$ is the only real solution to the system.