Algebraic Manipulation What is the best method to get the LHS equal to RHS? 
$\frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3}$
Thank you.
 A: Since $1=\frac 33$, one has $$\begin{align}\frac{n(n+1)(n+2)}{3}+\color{blue}{1}\cdot (n+1)(n+2)&=\frac{n(n+1)(n+2)}{3}+\frac{\color{blue}{3}(n+1)(n+2)}{\color{blue}3}\\&=\frac{n\color{red}{(n+1)(n+2)}}{\color{red}{3}}+\frac{3\color{red}{(n+1)(n+2)}}{\color{red}{3}}\\&=\color{red}{\frac{(n+1)(n+2)}{3}}(n+3)\end{align}$$
A: Your equation is :
$$ \frac{n(n+1)(n+2)}{3} + (n+1)(n+2) = \frac{(n+1)(n+2)(n+3)}{3} \tag{$\star$}$$
Notice that the product $(n+1)(n+2)$ is common to all the terms in $(\star)$. If $n \in \mathbb{Z} \smallsetminus \left\{-2,-1 \right\}$, then $(n+1)(n+2) \neq 0$ and you can $\color{blue}{\text{simplify}}$ by $(n+1)(n+2)$ as follows :
$$ \frac{\require{cancel} n\color{blue}{\cancel{(n+1)(n+2)}}}{3} + \require{cancel} \color{blue}{\cancel{(n+1)(n+2)}} = \frac{\require{cancel} \color{blue}{\cancel{(n+1)(n+2)}}(n+3)}{3} \tag{$\star$}$$
It gives :
$$ \frac{n}{3} + 1 = \frac{n+3}{3} = \frac{n}{3} + 1. $$
which is true for all $n \in \mathbb{Z} \smallsetminus \left\{-2,-1 \right\}$.
A: Firstly, take $(n+1)(n+2)$ out as a factor:
$((n+1)(n+2))[\frac n3 +1] = \frac{n+3}{3} (n+1)(n+2)$
Then divide through by $(n+1)(n+2)$:
$(\frac n3 +1) = \frac{n+3}{3}$
Multiply both sides by $3$:
$n+3 = n+3$
Therefore LHS = RHS
