Cubes from the Fifth Degree?

We observe that $f(1)=1$ , $f(2)=8$ and $f(3)=27$ , that is $f(n)=n^3$ for $n\in\{1,2,3\}$ . So, let us define another $5^{th}$ degree polynomial $g(x)$ such that $g(x)=f(x)-x^{3}$ .

Then, $x=1,2,3$ are 3 roots of the equation $g(x)=0$ . Since $f(x)$ is a $5^{th}$ degree polynomial in $x$ with real coefficients & leading coefficient unity, $g(x)$ will also be a $5^{th}$ degree polynomial in $x$ with real coefficients & leading coefficient unity, and we can rewrite g(x) as

$g(x)=(x-1)(x-2)(x-3)(x^2+ax+b)$ where $a,b \in R$

$\Rightarrow f(x)=(x-1)(x-2)(x-3)(x^2+ax+b)+x^3$

Now, it is also given that $f(0)=-6b=-6 \Rightarrow b=1$

Therefore, $f(4)-f(-1)$

$=[3 \times 2 \times 1 \times (16+4a+1)+64]$

$\;\;\;-[(-2) \times (-3) \times (-4) \times (1-a+1)+(-1)]$

$=6(17+4a)+64+24(2-a)+1$

$=102+24a+64+48-24a+1=215$