<s>[INST] Constructing a cubic given four points 
Question:  Is there an easier way to solve this problem?

Suppose the polynomial $f(x)$ is of degree $3$ and satisfies $f(3)=2$, $f(4)=4$, $f(5)=-3$, and $f(6)=8$.  Determine the value of $f(0)$.


My Attempt: I started off with the general cubic $ax^3+bx^2+cx+d=f(x)$ and manually plugged in each point to get the following system:$$\begin{align*} & 27a+9b+3c+d=2\\ & 64a+16b+4c+d=4\\ & 125a+25b+5c+d=-3\\ & 216a+36b+6c+d=8\end{align*}\tag1$$
Solving the system with the handy matrix gives the solutions as $a=\frac 92,b=-\frac {117}2,c=245,d=-328$. Thus, $f(0)=-328$.

Even though I (think) solved the problem correctly, this method seems a bit "bulky" especially when everything becomes a higher degree. So I'm wondering if there is a quicker way to evaluate this kind of problem.
 [/INST] You can directly find out the polynomial $f$ by considering it according as the $x$-values available: 
Let $f(x)=a_0+a_1(x-3)+a_2(x-3)(x-4)+a_3(x-3)(x-4)(x-5)$ for real constants $a_0,a_1,a_2,a_3$. Note that we don't need to take into account the value $x=6$ as this is already a cubic polynomial.
Then, $f(3)=2\Rightarrow a_0=2$
$f(4)=4\Rightarrow a_1=2$
$f(5)=-3\Rightarrow a_2=-\frac{9}{2}$
$f(6)=8\Rightarrow a_3=\frac{9}{2}$
Thus, $f(x)=2+2(x-3)-\frac{9}{2}(x-3)(x-4)+\frac{9}{2}(x-3)(x-4)(x-5)$.
I think the calculations are pretty simple this way as you have chosen $f$ to be such.
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