Does $x^2>x^3+1 \implies x < -{1\over P}$?

Question

How could one prove that:

$$x^2>x^3+1 \implies x < -{1\over P}$$

where $P$ denotes the plastic constant, the unique real root of $x^3-x-1=0$?

Answer 1

The function $x^2 - x^3 - 1$ is monotone decreasing on $\{x < 0\}$ so we need to look at the (unique) $x$ where $$x^2 = x^3 + 1.$$ But for this $x$, dividing by $x^3$, $$-\frac{1}{x} = -1 - \frac{1}{x^3},$$ so $-\frac{1}{x}$ solves the defining equation for the plastic constant; so $-\frac{1}{x} = P$ and $x = \frac{-1}{P}$.