<s>[INST] Prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$ I'm trying to prove that $n^3+3n^2+3n+1=2(1+2+...+n)(n+1)+(n+1)^2$, which is part of a larger proof. We can write:
$n^3+2n^2+n=2(1+2+...+n)(n+1)$
$n^3+2n^2+n=(1+2+...+n)(2n+2)$
$n^3+2n^2+n=(2n+4n+6n+...+2n^2)+(2+4+6+...+n)$
I have no idea how to proceed, though.
 [/INST] The sum $P(n):=1+2+\cdots n$ must be a quadratic polynomial in $n$, because $P(n)-P(n-1)=n$ is a linear polynomial.
Then as both members are cubic polynomials, it suffices to check equality for four distinct values of $n$.
$$\begin{align}n=0&\to 1^3=2()\cdot1+1^2&=1,\\
n=1&\to 2^3=2(1)\cdot2+2^2&=8,\\
n=2&\to 3^3=2(1+2)\cdot3+3^2&=27,\\
n=3&\to 4^3=2(1+2+3)\cdot4+4^2&=64.\end{align}
$$
This proves the identity for all $n$.

With the simplified identity (in my other answer), checking for three values suffices
$$\begin{align}n=0&\to 1^2=2()+1&=1,\\
n=1&\to 2^2=2(1)+2&=4,\\
n=2&\to 3^2=2(1+2)+3&=9.\end{align}
$$
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