= Solution
The finite sum of truncated cubes is $C^2$ by part (b) and is cubic on each <knot> <interval>. For $x\leq0$, every term vanishes. For $x\geq4h$, every truncated power equals its ordinary <polynomial>, so part (a) applied to $P(x)=x^3$ gives
$$
S_{(3)}(x)=\sum_{k=0}^4(-1)^k\binom4k(x-kh)^3=Q_{(4)}(x)=0.
$$
It is therefore zero on both exterior <intervals>. Its $C^2$ regularity makes the endpoint second <derivatives> zero too, proving naturality.
Evaluation at the five <knots> gives
$$
\boxed{\bigl(S_{(3)}(0),S_{(3)}(h),S_{(3)}(2h),S_{(3)}(3h),S_{(3)}(4h)\bigr)=h^3(0,1,4,1,0).}
$$
For example the three interior values in units of $h^3$ are $1$, $8-4=4$, and $27-4\cdot8+6=1$. At $4h$ the cancellation is $64-4\cdot27+6\cdot8-4=0$.
For $h=1$ an explicit graph description is
$$
S_{(3)}(x)=\begin{cases}
0,&x\leq0\text{ or }x\geq4,\\
x^3,&0\leq x\leq1,\\
4-6(x-2)^2+3|x-2|^3,&1\leq x\leq3,\\
(4-x)^3,&3\leq x\leq4.
\end{cases}
$$
It is symmetric about $x=2$, increases from zero to its maximum four at the <centre>, and decreases symmetrically back to zero. It joins the zero tails with zero first and second <derivatives>.
\Image[/past-exam-of-the-options-ii-course-of-the-university-of-oxford/2012-cardinal-cubic-spline.png]
{title=A compactly supported natural cubic spline with knot values zero, one, four, one, zero}
{description=The original plot uses the truncated-power formula for h equal to one. The maximum is four at x equal to two; both tails and the first two endpoint derivatives are zero.}
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