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?
 A: 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}
A: 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.
A: 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}
