Let $f(x)$ be a third-degree polynomial with real coefficients satisfying
\[|f(1)|=|f(2)|=|f(3)|=|f(5)|=|f(6)|=|f(7)|=12.\] Find $|f(0)|$.
Let $f(x)$ = $ax^3+bx^2+cx+d$.
Since $f(x)$ is a third degree polynomial, it can have at most two bends in it where it goes from up to down, or from down to up.
By drawing a coordinate axis, and two lines representing $12$ and $-12$, it is easy to see that $f(1)=f(5)=f(6)$, and $f(2)=f(3)=f(7)$; otherwise more bends would be required in the graph. Since only the absolute value of $f(0)$ is required, there is no loss of generalization by stating that $f(1)=12$, and $f(2)=-12$. This provides the following system of equations.
\[a +     b +   c +   d =  12\]
\[8a +   4b + 2c +   d = -12\]
\[27a +   9b + 3c +   d = -12\]
\[125a + 25b + 5c +   d =  12\]
\[216a + 36b + 6c +   d =  12\]
\[343a + 49b + 7c +   d = -12\]
Using any four of these functions as a system of equations yields $d = |f(0)| = \boxed{072}$