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 $\frac{4}{27}=\max_{x \in [0,1]}x(1-x)^{2}$.
 A: Define $F(x)= x(1-x)^2$. Then
$F(0)=F(1)=0$, and $$F(\frac{1}{3})=\frac{4}{27}=\max_{x \in [0,1]}F(x).$$
Since $0<s < \frac{4}{27}$, by the Intermediate Value Theorem, the equation $F(x)=s$ admits at least one solution in each of the intervals $(0,\frac{1}{3})$ $(\frac{1}{3},1)$.
It remains to show that each interval contains at most one solution. This follows from monotonicity:
$$
F'(x)=3x^2 - 4x + 1>0 \iff x < \frac{1}{3} \,\,\text{ or }\,\, x>1.
$$
Thus $F|_{(0,\frac{1}{3})}$ is decreasing, while $F|_{(\frac{1}{3},1)}$ is increasing.

For any $s>0$, since $F(1)=0$, and $\lim_{x \to \infty} F(x)=\infty$, it follows that there exists $y>1$ such that $F(y)=s$.
A: 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.
