24, 18, 24, 18, will this continue?

Let $f(x)=ax^3+bx^2+cx+d$

So $f(1)=a+b+c+d=24$

$f(2)=8a+4b+2c+d=18$

$f(3)=27a+9b+3c+d=24$

$f(4)=64a+16b+4c+d=18$

If you solve this system of equations above, you will get that the only possible values for $a,b,c,d$ are

$a=-4,b=30,c=-68,d=66$

So the only possible cubic polynomial is $f(x)=-4x^3+30x^2-68x+66$

So $f(5)=-500+750-340+66=-24.$

For a cubic, $f(x)$ , the second differences, $g(x)=2f(x)-f(x+1)-f(x-1)$ , form a linear function. Now $g(2)=2 \times 18-24-24=-12$ and $g(3)=12$ so $g(4)=36-24-f(5)=36$ and $f(5)=\boxed{-24}$

We note that the following cubic polynomial satisfies the system of equations.

$\begin{aligned} f(x) & = {\color{#3D99F6}-4(x-2)(x-3)(x-4)} + 9(x-1)(x-3)(x-4) - 12(x-1)(x-2)(x-4) + 3(x-1)(x-2)(x-3) \end{aligned}$

Note that the cubic sub-polynomial $\color{#3D99F6} -4(x-2)(x-3)(x-4) = \begin{cases} 24 & \text{when }x = 1 \\ 0 & \text{when }x=2,3,4 \end{cases}$ . Similarly for the other three cubic sub-polynomials, so that $f(x) = \begin{cases} 24 & \text{when }x = 1, 3 \\ 18 & \text{when }x = 2, 4 \end{cases}$

Therefore,

$\begin{aligned} f(5) & = -4(3)(2)(1) + 9(4)(2)(1) - 12(4)(3)(1) + 3(4)(3)(2) \\ & = -24 + 72 - 144 + 72 \\ & = \boxed{-24} \end{aligned}$