Show that there is at least one $\varepsilon\in(0,1)$, such that $ f(x)=-1+x^2+\frac{x^2(x-1)}{3!}f'''(\varepsilon),\quad x\in(0,1). $ 
Suppose $f(x)$ is three-times differentiable on $[0,1]$, and $f(0)=-1,\;f(1)=0,\;f'(0)=0$. Show that for any $x\in(0,1)$ there is at least one $\varepsilon\in(0,\,1)$, such that
  $$
f(x)=-1+x^2+\frac{x^2(x-1)}{3!}f'''(\varepsilon).
$$

The Taylor expansion of $f$ at $0$ is 
$$
f(x)=-1+\frac{1}{2}f''(0)x^2+\frac{x^3}{3!}f'''(\varepsilon_1),
$$
and at $1$ is
$$
f(x)=f'(1)(x-1)+\frac{1}{2}f''(1)(x-1)^2+\frac{(x-1)^3}{3!}f'''(\varepsilon_2).
$$
But I don't know what I can get.
 A: The map $f_1(x)= f(x) -(-1+x^2)$ is satisfying $f_1(0)=f_1(1)=f_1^\prime(0)=0$.
Take $x \in (0,1)$ and consider the function
$$g(t) = f_1(t) -A t^2(t-1)$$ where $A$ is chosen such that $g(x)=0$, i.e. $A =\frac{f_1(x)}{x^2(x-1)}$.
As $g(0)=g(1)=g(x)=0$, according to Rolle's theorem it exists $\alpha_1 \in (0,x)$ and $\alpha_2 \in (x,1)$ such that $g^\prime(\alpha_1) = g^\prime(\alpha_2) =0$. We also have $g^\prime(0)=0$. Therefore one can find $\beta_1 \in (0,\alpha_1)$ and $\beta_2 \in (\alpha_1, \alpha_2)$ with $g^{\prime\prime}(\beta_1)=g^{\prime\prime}(\beta_2) = 0$.
According to Rolle's theorem again, it exists $\epsilon \in (\beta_1,\beta_2) \subseteq (0,1)$ such that $g^{\prime\prime\prime}(\epsilon) = 0$, i.e. $g^{\prime\prime\prime}(\epsilon) = f_1^{\prime\prime\prime}(\epsilon)-3!A =f^{\prime\prime\prime}(\epsilon)-3!A= 0$.
Finally $A =\frac{f_1(x)}{x^2(x-1)} = \frac{f^{\prime\prime\prime}(\epsilon)}{3!}$ leading to the expected result
$$f(x)=-1+x^2+\frac{x^2(x-1)}{3!}f'''(\varepsilon)$$
