To find a polynomial My doubt is from a PRMO model paper I had today. The image of the question is given below ( I had to have my lunch, so thought of a quicker way to put my question and ended up with this) :

What I tried :
I felt that $f(x) - x^3$ can give me the value of the quadratic part of the polynomial.
As a result, taking the quadratic part to be of the form $ax^2 +bx + c$, the differences I get are :

*

*$f(1) - 1^3 = a + b + c = 0$

*$f(2) - 2^3 = 4a + 2b + c = -4$

*$f(3) - 3^3 = 9a + 3b + c = -18$
I am not an expert at solving 3 linear equations in 3 variables , but I tried and ended up pulling out my hair (trying to be a bit literary; hope you won't mind the wordings , but rather concentrate on the question). I tried to take 2 equations at a time, and ended getting multiple values for the same variables.
I will be grateful to anyone who is willing to help me out.
 A: Hint: Consider $f(x)-x^2$ instead. It's so much easier to work with polynomials that are 0 at given points.
A: While the solutions involving $x^2$ are likely the ones intended by the question-setter, it is also easy to quickly get the value of $f(4)$ without extracting the quadratic by using finite differences.  Let $g(n) = f(n) - n^3$, as in the OP, be quadratic.  The first-order differences are:
$$\Delta g(1) = g(2)-g(1) = -4 \\
\Delta g(2) = g(3) - g(2) = -14.$$
So the second-order difference is $\Delta^2 g(1) = \Delta g(2) - \Delta g(1) = -10$, which for any quadratic (or lower) polynomial is constant.  Hence $\Delta^2 g(2) = -10$, so $\Delta g(3) = \Delta g(2) -10 = -24$, and $g(4) = g(3) -24 = -42$.
So $f(4) = 4^3 -42 = 22$.
A: Note that $f(x) = x^2$ satisfy the three given equations. But we want of degree $3$ so we add $(x-1)(x-2)(x-3)$ and get $$f(x) = (x-1)(x-2)(x-3) + x^2$$ with $f(4) = 6+16 = 22$.
A: We don't need to determine explicitly the coefficients for the polynomial, indeed we have that by uniqueness
$$f(x)=(x-1)(x-2)(x-3)+\frac12(x-2)(x-3)-4(x-1)(x-3)+\frac92(x-1)(x-2)$$
which satisfies by construction the given conditions with $f(1)=1$, $f(2)=4$ and $f(3)=9$, then
$$f(4)=(3)(2)(1)+\frac12(2)(1)-4(3)(1)+\frac92(3)(2)=6+1-12+27=22$$
As noticed by other answers, more trickly we have that
$$f(x)=(x-1)(x-2)(x-3)+x^2$$
A: We have $$a+b+c=0,$$ $$4a+2b+c=-4$$ and $$9a+3b+c=-18,$$ which gives $$(a,b,c)=(-5,11,-6)$$ and $$f(x)=x^3-5x^2+11x-6.$$
Thus, $$f(4)=22.$$
A: *

*$f(1) - 1^3 = a + b + c = 0$

*$f(2) - 2^3 = 4a + 2b + c = -4$

*$f(3) - 3^3 = 9a + 3b + c = -18$
The third equation is wrong
Subtracting eq(1) from eq(2)
$ 4a + 2b + c   -(  a  + b + c) =  -4-0$
$3a +b = -4$------------------------------                       eq(3)
Subtracting eq(1) from eq(3)
$9a + 3b + c -(4a + 3b + c) = -18-(-4)$
$5a = -22$
$a = \frac{-2}{5}$
Substitute this in all equations and you'll get the answers
A: Given $f(x)$ , a monic cubic polynomial.
$f(1) = 1$, $f(2) = 4$, $f(3) = 9$
By factor theorem we can show that $(x-1)$,$(x-2)$ and $(x-3)$ are factors of $f(x) - x^2$
We have three linear factors thus we can write $f(x)$ as,
$f(x) - x^2 = k(x-1)(x-2)(x-3)$,
where k is some constant
But $f(x)$ is a monic polynomial, therefore k = 1
thus our polynomial becomes,
$f(x) - x^2 = (x-1)(x-2)(x-3)$
Putting x = 4,
$f(4) = 6 + 16 = 22$
This is common technique to solve competition math problems,
You can read more about it here
https://brilliant.org/wiki/polynomial-interpolation/
