# How do you solve (m^2 + 2m + 1)/( m^3 + 3m^2 + 3m +1)(m^2/(m - (3m)/3)) ?

Mar 11, 2018

The function doesn't exist.

#### Explanation:

$\textcolor{w h i t e}{=} \frac{{m}^{2} + 2 m + 1}{{m}^{3} + 3 {m}^{2} + 3 m 1} \left({m}^{2} / \left(m - \frac{3 m}{3}\right)\right)$

$= \frac{{m}^{2} + 2 m + 1}{{m}^{3} + 3 {m}^{2} + 3 m 1} \left({m}^{2} / \left(m - \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}} m}{\textcolor{red}{\cancel{\textcolor{b l a c k}{3}}}}\right)\right)$

$= \frac{{m}^{2} + 2 m + 1}{{m}^{3} + 3 {m}^{2} + 3 m 1} \left({m}^{2} / \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{m - m}}}\right)\right)$

$= \frac{{m}^{2} + 2 m + 1}{{m}^{3} + 3 {m}^{2} + 3 m 1} \left({m}^{2} / 0\right)$

Since there is division by zero, the function cannot exist.