Polynomial : $ P(x+1)-2P(x)+P(x-1)=6x $ Find all polynomials $P(x) \in \mathbb{R}[x]$ satisfying 
$$ P(x+1)-2P(x)+P(x-1)=6x $$
My attempt :
Since $ P(x+1)+P(x-1)-2P(x)=6x $, so $P(x)$ is not constant polynomial.
Let $P(x+1)-P(x)= Q(x)$
so $Q(x)-Q(x-1)=6x, \;\; \forall x \in \mathbb{R}$
by induction,  $Q(x+n)-Q(x)= 6((x+n)+(x+n-1)+...+(x+1))$
Continue from dxiv's answer,
$Q(n)-Q(0) = 6(n+(n-1)+...+1)$
$Q(n)=Q(0) + 6(n+(n-1)+...+1)=Q(0) + 3n(n+1) $
so $P(n) = Q(n-1) + Q(n-2)+ ...+Q(0)+P(0)$
$= \displaystyle\sum^{n-1}_{i=1}3i(i+1) + nQ(0)+P(0)$
$= \displaystyle\sum^{n-1}_{i=1}3i(i+1) + nP(1)- nP(0)+P(0)$
$= \displaystyle\sum^{n-1}_{i=1}3i(i+1) + nP(1)-(n-1)P(0)$
$= 3\displaystyle\sum^{n-1}_{i=1}(i^2+i) + nP(1)-(n-1)P(0)$
$= n^3-n + nP(1)-(n-1)P(0)\;\; \forall x \in \mathbb{N}$
Since $P(x)-x^3-x(P(1)-1)+(n-1)P(0)$ has infinitely many roots so 
$P(x)-x^3-x(P(1)-1)+(n-1)P(0) = 0$, we get
$P(x)=x^3+x(P(1)-1)-(n-1)P(0)\;\; \forall x \in \mathbb{R}$
As $P(1)-1$ and $(n-1)P(0)$ are in $\mathbb{R}$, we obtain
$P(x)=x^3+cx+d,\; \forall x \in \mathbb{R}$ and $c,d$ are real constants.
 A: Let $\delta$ be the operator mapping a polynomial $p(x)$ into the polynomial $(\delta p)(x)=p(x+1)-p(x)$.


*

*If $p$ is non-constant, the degree of $\delta p$ is the degree of $p$ minus one;

*If the leading term of $p(x)$ is $ax^n$ with $n\geq 1$, the leading term of $(\delta p)(x)$ is $na x^{n-1}$;

*If $p(x)=\binom{x}{k}$ with $k\geq 1$, then $(\delta p)(x)=\binom{x}{k-1}$;

*Any polynomial can be represented in the binomial base and with such representation the operator $\delta$ essentially acts as a shift by the previous point.


The given problem can be stated as
$$ (\delta^2 P)(x) = 6\binom{x+1}{1} $$
hence a solution is given by
$$ P(x) = 6\binom{x+1}{3} = \color{blue}{x^3-x}.$$
$(\delta^2 P)(x)$ equals zero iff the degree of $P$ is $\leq 1$, hence the full set of solutions is given by $\color{blue}{x^3+ax+b}$.
A: 
Let $\;P(x+1)-P(x)= Q(x)\;$ so $\;Q(x)-Q(x-1)=6x$
by induction,  $Q(x+n)-Q(x)= 6((x+n)+(x+n-1)+...+(x+1))$

For $\,x=0\,$: $\displaystyle\;\;Q(n)=Q(0) + 6\big(n+(n-1)+\cdots+1\big)=Q(0)+3n(n+1)\,$, then:
$$\,P(n)=Q(n-1)+Q(n-2)+\cdots+Q(0)+P(0)= P(0)+n\,Q(0)+ 3\sum_{k=0}^{n-1} k(k+1)=\cdots$$
A: Hint:
WLOG $$P(x)=\sum_{r=0}^na_rx^r$$
Compare the constants & the coefficient of the increasing powers of $x$ to find $a_r$s
A: Note if $P$ and $Q$ are two different polynomials which satisfy the given relation, then $R = P-Q$ satisfies $$R(x+1) - 2R(x) + R(x+1) = 0$$ which implies $R$ is linear: we get that $R(n) = an + b$ forms an arithmetic progression for integer $n$, and hence $R(x) = ax + b$ for all $x$. 
The last statement follows from the fact that if two polynomials take the same values at infinite points, then they are identical.
One can see that $P(x) = x^3$ is a solution (using binomial theorem) and so the general form of the solution is
$$P(x) = x^3 + ax + b$$
It is easy to verify that for any $a, b$ this satisfies the original equation.
A: I did in a pretty simpler way than you did.
$p(x)=ax^2+bx+c$ doesn't work because $p(x-1)+p(x+1)-2 p(x)=2a$ is a constant
so I thought to $3rd$ degree $p(x)=ax^3+bx^2+cx+d$
computed $p(x-1)+p(x+1)-2 p(x)=2 b + 6 a x$ and concluded that it must be $a=1;\;b=0$
So all third degree polynomial which satisfy the request are of the form $p(x)=x^3+cx+d$ for any $c,\;d\in\mathbb{R}$
Fourth degree or higher like $p(x)=a x^4+b x^3+c x^2+d x+e$ can't satisfy the request because $p(x-1)+p(x+1)-2 p(x)=12 a x^2+2 a+6 b x+2 c$ and the condition to eliminate $x^2$ or other powers greater than is to put $a=0$ and the polynomial is no more $4-$th degree.
A: It should be $P(x)=ax^3+bx^2+cx+d$.
The condition gives $6ax+2b=6x$.
Thus, $a=1$, $b=0$ and $P(x)=x^3+cx+d$.
$$P(x)=P(0)+P'(0)x+\frac{P'(0)}{2!}x^2+...+\frac{P^{(n)}(0)}{n!}x^n=$$
$$=P(1)+P'(1)(x-1)+\frac{P'(1)}{2!}(x-1)^2+...+\frac{P^{(n)}(1)}{n!}(x-1)^n=$$
$$=P(-1)+P'(1)(x+1)+\frac{P'(-1)}{2!}(x+1)^2+...+\frac{P^{(n)}(-1)}{n!}(x+1)^n,$$
which gives
$$P(x+1)=P(1)+P'(1)x+\frac{P'(1)}{2!}x^2+...+\frac{P^{(n)}(1)}{n!}x^n$$ and
$$P(x-1)=P(-1)+P'(-1)x+\frac{P'(-1)}{2!}x^2+...+\frac{P^{(n)}(-1)}{n!}x^n.$$
Thus,
$$6x=P(x+1)-2P(x)+P(x+1)=$$
$$=P(1)-2P(0)+P(-1)+\left(P'(1)-2P'(0)+P'(-1)\right)x+$$
$$+\frac{P''(1)-2P''(0)+P''(-1)}{2!}x^2+...\frac{P^{(n)}(1)-2P^{(n)}(0)+P^{(n)}(-1)}{n!}x^n$$
and since ... ?
A: If we take derivative of this equation twice, we get $$P''(x+1)-2P''(x)+P''(x-1)=0$$
So if we put $a_n = P''(n)$ we get $a_n ={a_{n-1}+a_{n+1}\over 2}$ which means that $a_n$ is arithmetic sequence, thus $a_n = an+b$ for some $a,b$. But then $P''(x) = ax+b$ for all $x$, and thus $P(x) = a_1x^3+b_1x^2+c_1x+d_1$ for some $a_1,b_1,c_1,d_1$. Now plug this in to original equation and calculate $a_1, b_1,c_1,d_1$.
