Let's do some calculus! (9)

Rewriting as $\displaystyle\lim_{x\to a}\frac{F(x)-F(a)}{(x-a)^3} - \lim_{x\to a}\frac{f(x)+f(a)}{2(x-a)^2}$ and applying L'Hospital's Rule twice results in

$\displaystyle\lim_{x\to a}\frac{f'(x)}{6(x-a)} - \lim_{x\to a}\frac{f''(x)}{4}$

Now apply once to the first limit to make $\displaystyle\lim_{x\to a}\frac{f''(x)}{6} - \lim_{x\to a}\frac{f''(x)}{4}$

Now it is possible to apply direct substitution for x = a, so $\displaystyle\frac{-f''(a)}{12}$ must be zero for all a. This is only possible if $f''(x) =$ a constant, making $f(x)$ a function of degree $\boxed{1}$

Note:

If $f(x)$ were to be a higher degree function, $\frac{-f''(a)}{12}$ may not be zero making the original equation false, thus eliminating all degrees higher than 1.