Find a cubic polynomial. If $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\ f(6/5)=(6/5)^3?$
I attempt:
I managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting in the equation and so on....
But we can see:
$f(x)=q_1(x-1)+1\\f(x)=q_2(x-2)+4\\f(x)=q_3(x-3)+9$ 
Also when we put $x=1$ we get $f(1)=1^2$, when $x=2$ then $f(2)=2^2$, when $x=3$ then $f(3)=3^2$ but $f(4)\neq4^2$ (from answer).
Can this information be used to reproduce $f(x)$ directly without using the step I described in very first line of my solution?

If 


*

*$f(x) = x^3 + ax^2 + bx + c$

*$f(1) = 1$

*$f(2) = 4$

*$f(3) = 9$
then
\begin{align}
   f(x+1) - f(x) &= (3x^2+3x+1) + a(2x+1) + b \\
   \hline
   4-1 &= f(2)-f(1) \\
   3 &= 7 + 3a + b \\
\hline
   9-4 &= f(3) - f(2) \\
   5 &= 19 + 5a + b \\
\hline
   3a + b &= -4 \\
   5a + b &= -14 \\
\hline
   a &= -5 \\
   b &= 11 \\
   c &= -6
\end{align}
S0 $f(x) = x^3 - 5x^2 + 11x - 6$.

You could make a difference table
\begin{array}{c}
   1 && 4 && 9 \\
   & 3 && 5 \\
   && 2
\end{array}
Using $f(x) = x^3 + ax^2 + bx + c$, this corresponds to
\begin{array}{c}
   1+a+b+c && 8 + 4a + 2b + c && 27 + 9a + 3b + c \\
   & 7+3a+b && 19 + 5a + b \\
   && 12+2a
\end{array}
Then, comparing entries...
\begin{align}
   12+2a=2 &\implies a = -5 \\
   7+3a+b = 3 &\implies b=11 \\
   1+a+b+c = 1 &\implies c = -6
\end{align}
