Searching for Polynomials!

Given: $\left(x^3 + 3x^2 + 3x + 2 \right) P(x-1) =\left(x^3 - 3x^2 + 3x - 2 \right)P(x)$

$\displaystyle \implies (x+2)(x^2+x+1)P(x-1)=(x-2)(x^2-x+1)P(x)$

$\displaystyle \implies \dfrac{P(x)}{P(x-1)}=\dfrac{(x+2)(x^2+x+1)}{(x-2)(x^2-x+1)}$

$\displaystyle \implies \prod_{x=3}^{n} \dfrac{P(x)}{P(x-1)} = \prod_{x=3}^{n} \dfrac{(x+2)(x^2+x+1)}{(x-2)(x^2-x+1)}$

$\displaystyle \implies \prod_{x=3}^{n} \dfrac{P(x)}{P(x-1)} = \prod_{x=3}^{n} \dfrac{(x+2)\cdot T_{x}}{(x-2)\cdot T_{x-1}} \ (*)$

where $T_{x}=x^2+x+1$

Clearly, $(*)$ is a telescoping product, evaluating it, we have,

$\displaystyle\dfrac{P(n)}{P(2)} = \dfrac{1}{4!} \dfrac{(n+2)!(n^2+n+1)}{7(n-2)!}$

$\displaystyle \implies P(n) = \dfrac{n(n+2)(n^2-1)(n^2+n+1)}{2}$

$\therefore P(10)=659340$

In what follows, we are going to use the following notation $w_{1}=e^{\frac{\pi}{3}i}$ and $w_{2}=e^{\frac{2 \pi}{3}i}.$ Now it is not difficult to see that the given equation can be written in the following way $(x+2)(x-w_{2})(x- \overline {w_{2}}) P(x-1)=(x-2)(x-w_{1})(x-\overline {w_{1}})P(x)$ Therefore the numbers $-2, w_{2}, \overline{w_{2}}$ are roots of $P(x)$ , and the numbers $2, w_{1}, \overline {w_{1}}$ are roots of $P(x-1)$ . Using that if a number $\alpha$ is a root of a polynomial $P(x-1)$ , then $\alpha-1$ is a root of $P(x),$ we obtain three other roots of $P(x)$ by subtracting $1$ from each number $2, w_{1},$ or $\overline {w_{1}},$ respectively. Those roots will be $1, w_{1}-1=w_{2}, \overline {w_{1}}-1=\overline {w_{2}},$ but only $1$ is a new root because we already know that $w_{2}$ and $\overline{w_{2}}$ were roots of $P(x).$ So a partial conclusion here would be that $P(x)$ has the roots $1, -2, w_{2}, \overline{w_{2}}.$ Then $P(x)=(x-1)(x+2)(x-w_{2})(x-\overline {w_{2}})Q(x)$ where $Q(x)$ is a polynomial.

By substituting $P(x)$ by the latter expression in the given equation of the problem and cancelling the factors that appear in both sides, we obtain the equation $(x+1)Q(x-1)=(x-1)Q(x) \:\: (*).$

From this equations and using a reasoning similar to the one that we used before, we obtain that $Q(x)$ has the roots $-1$ and $0$ and therefore can be written like $Q(x)=x(x+1)R(x)$ where $R(x)$ is a polynomial. By substituting in $(*)$ and simplifying again, we get $R(x-1)=R(x)$ . The latter equation implies that $R(x)=R(x+n)$ for any integer $n$ , and this implies that $R(x)$ is a constant polynomial. Denoting this constant by $c$ we obtain that $P(x)=c x(x+1)(x-1)(x+2)(x-w_{2})(x-\overline w_{2}).$ Using the condition $P(2)=84$ we get that $c=\frac{1}{2}$ , and therefore $P(10)=(1/2) (10)(11)(9)(12)(10^2+10+1)=659340.$