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.

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 $n \in \mathbb{Z} \smallsetminus \left\{-2,-1 \right\}$.
