How can you prove $\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$ without much effort? 
I will keep it short and take only an extract (most important part) of
  the old task.
$$\frac{n(n+1)(2n+1)}{6}+(n+1)^2= \frac{(n+1)(n+2)(2n+3)}{6}$$

What I have done is a lot work and time consuming, I have "simply" solved it. But I think with a lot less work, there would be an easier and faster way. It's just I cannot see it : /
If anyone wants see, here is my long solution which I'm not happy with:
$$\frac{n(n+1)(2n+1)+6(n+1)^2}{6}=\frac{(n^2+2n+n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{(2n^3+n^2+2n^2+n)+6n^2+12n+6}{6} = \frac{(n^2+3n+2)(2n+3)}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+3n^2+n+6n^2+12n+6}{6}=\frac{2n^3+3n^2+6n^2+9n+4n+6}{6} \Leftrightarrow$$
$$\Leftrightarrow \frac{2n^3+9n^2+13n+6}{6}=\frac{2n^3+9n^2+13n+6}{6}$$
 A: Here is one solution. Since each side of identity is a polynomial of third degree (you can reduce it to degree 2 by canceling $(n+1)$), you can just verify that it holds for $4$ (or just $3$ after cancellation) distinct values of $n$ say $n=1,0,-1,-2$ then you are done. Note that I have chosen specific values of $n$ in such a manner to make calculations easy (some terms vanish for $n=0,-1,-2$).
This is useful if expressions are complicated. See the end of this answer https://math.stackexchange.com/a/1814894/72031 where I use this technique to prove an identity involving polynomials of degree $4$.
Update: this is identical to Hagen von Eitzen's comment which I saw later. I am marking it as community wiki. 
A: Note that
$$n(2n+1)+6(n+1)=2n^2+7n+6=(n+2)(2n+3)$$
Multiplying both sides by $\dfrac {n+1}6$
$$\frac {n(n+1)(2n+1)}6+(n+1)^2=\frac {(n+1)(n+2)(2n+3)}6$$
A: Since you want $n+1$ to appear as a factor at the end, I would just leave that as a factor:
$$
\frac{n(n+1)(2n+1)+6(n+1)^2}6 = \frac {n(2n+1) + 6(n+1)} 6 (n+1).
$$
That becomes
$$
\frac {2n^2 + n + 6n + 6} 6 (n+1) = \frac{2n^2 + 7n+6} 6 (n+1) = \frac{(n+2)(2n+3)} 6 (n+1)
$$
A: Oh goodness, you needn't so much work.  All you need to do is this:
$$\begin{align}\frac{n(n+1)(2n+1)}6+(n+1)^2&=\frac{n(n+1)(2n+1)}6+\frac{6(n+1)^2}6\\&\tag 1=c\bigg(n(2n+1)+6(n+1)\bigg)\\&\tag2=c\bigg(2n^2+7n+6\bigg)\\&\tag3=c\bigg((n+2)(2n+3)\bigg)\end{align}$$
$(1)$ factor out $(n+1)/6$ and call it $c$.
$(2)$ expand the remaining stuff
$(3)$ factor the remaining stuff.
A: The most general method: do all multiplications on both sides to transform both numerators to canonical form, then verify polynomials are the same degree and have respective coefficients equal.
