If $P(x)$ is a unique cubic polynomial for which $P(x_0)=f(x_0),P(x_2)=f(x_2),P^{'}(x_1)=f^{'}(x_1),P^{''}(x_1)=f^{''}(x_1)$,$f(x)$ is a given function differentiable $4$ times.
Show that $f(x)-P(x)=\dfrac{x^4-1}{4!}f^{4}(c) $ where $x_0=-1,x_1=0,x_2=1,c\in (-1,1) $
MY TRY:
Let us assume that $P(x)=a_0+a_1x+a_2x^2+a_3x^3.$
Then $$\begin{aligned}P(-1)=a_0-a_1+a_2-a_3&=f(-1)\quad &(1)\\
P(1)=a_0+a_1+a_2+a_3&=f(1) \quad &(2)\\
P^{'}(0)=a_1&=f^{'}(0)\quad &(3)\\
P^{''}(0)=2a_2&=f^{''}(0)\quad &(4)
\end{aligned}$$
How can I prove the given fact using the $4$ relations I have found here?
Please help.