Factor out (m+1) in the following so that the final answer is $\frac{(2m+1) (m+2) (m+1)} {6}$ Question: $\frac{m (m+1) (2m+1) + 6(m+1)^2}{6}$=$\frac{(2m+3)(m+2)(m+1)}{6}$
I must multiply by 6 on both sides and expand the brackets and collect like terms. I'm I correct?
Edit notes: The original problem was posed as:
$$ \frac{(m+1) (2m+1) + 6(m+1)^2}{6}= \frac{(2m+3)+(m+2)(m+1)}{6} $$
which did not match the title. The edit provides the corrected question. 
 A: The expression in the title can't be the simplified answer, because as a test, if we substitute $m=0$ in your original (you could use any number but $0$ is convenient) we get
$$\dfrac {(2(0)+1)(0+1)+6(0+1)^2}{6}=\dfrac 76$$
Substituting $m=0$ in the expression in the title yields
$$\dfrac {(2(0)+1)(0+2)(0+1)}{6} = \dfrac 26$$
So the $2$ expressions cannot be equal. What you can do to simplify your original problem is the following: 
Both terms in the numerator, $(2m+1)(m+1)$ and $6(m+1)^2$, have a common factor of $(m+1)$, so we can factor that out
$$\dfrac {(m+1)[(2m+1)+6(m+1)]}{6}$$
Now you should distribute the $6$ in the numerator and combine like terms
$$\dfrac {(m+1)(2m+1+6m+6)}{6}$$
$$\dfrac {(m+1)(8m+7)}{6}$$
If you want to keep all integers, this is as far as you can go.
A: The description as given is correct. Multiplying both sides by $6$ yields
$$(2m+1) (m+1) + 6(m+1)^2= (2m+3)+(m+2)(m+1)$$
Now what remains is to determine the value of both sides. Let $L_{m}$ be the left and $R_{m}$ be the right. 
\begin{align}
L_{m} &= (2m+1)(m+1)m + 6(m+1)^2 \\
&= (2m^2 + 3m + 1)m + 6(m^2 + 2m +1) \\
&= 2 m^3 + 9 m^2 + 13 m + 6. 
\end{align}
\begin{align}
R_{m} &= (2m+3)(m+2)(m+1) \\
&= 2 m^3 + 9m^2 + 13 m + 6. 
\end{align}
This demonstrates that $L_{m} = R_{m}$, as required.
