Find a cubic polynomial.

Consider $g(x) = f(x)-x^2$. Then $g(1) = g(2) = g(3) = 0$ and $g$ is also a cubic polynomial and has leading coefficient 1. Thus $g(x) = (x-1)(x-2)(x-3)$ and hence $f(x) = (x-1)(x-2)(x-3)+x^2$. It now follows that $f(4) = 22$. Other values can be calculated.

Classic long method:

Let $f(x) = x^3 + b x^2 + c x + d$ with $f(1) = 1$, $f(2) = 4$, $f(3) = 9$, which leads to \begin{align} f(1) &= 1 = 1 + b + c + d \hspace{10mm} \to d = -b - c \\ f(2) &= 4 = 8 + 4 b + 2 c + d = 8 + 3b + c \hspace{10mm} \to c = -4 - 3b, \, d = 4 + 2b \\ f(3) &= 9 = 27 + 9b + 3c + d = 19 + 2b \end{align} from which $b = -5$, $c = 11$, and $d = -6$ and $$f(x) = x^3 - 5 \, x^2 + 11 \, x -6.$$

With $f(x)$ then \begin{align} f(4) &= 64 - 80 + 55 -6 = 22 \\ f\left(\frac{6}{5}\right) &= \left(\frac{6}{5}\right)^{3} - \frac{36 - 66 + 30}{5} = \left(\frac{6}{5}\right)^{3}. \end{align}

It may also be noticed that $f(x)$ can be seen in the form $$f(x) = \left(x - \frac{5}{3}\right)^{3} + \frac{8}{3} \, \left( x - \frac{5}{3}\right) + \frac{83}{27}.$$ From this it is easy to see that \begin{align} f\left(\frac{5}{3}\right) &= 3 + \frac{2}{27} \\ f\left(\frac{5}{6}\right) &= \frac{59}{216}. \end{align}