hmm

Here's my attempt:

Let the function be in the form \(ax^3 + bx^2 + cx + d = y\). Also note that \(f(1) = 9\) means that \((1,9)\) is a point on this graph. Subbing this in gives us \(a + b + c + d = 9\)

With stars and bars, we use 3 bars, so it is \({9 + 3 \choose 3} = 220\). But, this accounts for \(a = 0\), which we don't want; it has to be a cubic polynomial. 

If we "remove a", there are 9 stars, but only 2 boxes, so there are \({9 + 2 \choose 2} = 55\) ways for \(a = 0\).

So, there are \(220 - 55 = \color{brown}\boxed{165}\) polynomials.