How many roots of $x(1-x)^{2}=s$ are there in $(0,1)$? This is a self-answered question, which is part of answering this related question. Alternative solutions are welcomed.
Let $0<s < \frac{4}{27}$. Prove that the cubic equation $x(1-x)^{2}=s$ has exactly two solutions in $(0,1)$.
Moreover, the third solution is a real number greater than $1$.
The limitation on the range of $s$ is due to $\frac{4}{27}=\max_{x \in [0,1]}x(1-x)^{2}$.

Define $F(x)= x(1-x)^2.$ Then, $F'(x)=(1-x)(1-3x)$ hence:

*

*for all $x<0,$ $F(x)<F(0)=0;$

*as $x$ increases from $0$ to $1/3,$ $F(x)$ continuously increases from $0$ to $F(1/3)=4/27;$

*as $x$ increases from $1/3$ to $1,$ $F(x)$ continuously decreases from $4/27$ to $F(1)=0;$

*as $x$ increases from $1$ to $+\infty,$ $F(x)$ continuously increases from $0$ to $\lim_{+\infty}F=+\infty.$
This ends the proof.
