Summer Barbecubics

Let $f(x) = ax^3+bx^2+cx+d$ . Substitute the $4$ given coordinates into $f(x)$ , and you will get this system of equations:

$\begin{cases} f(-1)=3&-a+b-c+d=3\\ f(-2)=-12&-8a+4b-2c+d = -12 \\ f(0)=6&0a+0b+0c+d = 6 &\implies d=6\\ f(2)=24&8a+4b+2c+d=24 \end{cases}$

Substitute the value of $d$ into the remaining $3$ equations, and you should have

$\begin{cases} -a+b-c=-3&\implies\boxed{1}\\ -8a+4b-2c=-18 &\implies\boxed{2}\\ 8a+4b+2c=18 &\implies \boxed{3}\end{cases}$

$\boxed{2}+\boxed{3}$ :

$(-8a+4b-2c)+(8a+4b+2c)=-18+18\\ 8b=0\\ b=0$

Substitute the value of $b$ into $\boxed{1}$

$-a+0-c = -3 \implies a=3-c \implies\boxed{4}$

Substitute the value of $b$ and $\boxed{4}$ into $\boxed{3}$ :

$8(3-c)+4(0)+2c = 18\\ 24-8c+2c=18\\ 6c=6\\ c=1\\ \implies a=3-1=2$

The function is $f(x) = 2x^3+x+6$ . Therefore,

$f(f(1))\\ =f(2(1)^3+1+6)\\ =f(9)\\ =2(9)^3+9+6\\ =\boxed{1473}$