1. ## Simplifying an expression

Hello,
I would like to know how to simplify this expression step-by-step:
Wolfram Alpha

If I try and do it by hand on paper, I get results like this:
(I forgot to add bottom of fraction at the end, because top was already wrong, but the point of the problem remains...)

2. Don't expand out until you have factorized as fully as possible.

In this case, we have
$\displaystyle \frac{(m+n)(m^2-n^2)-(m-n)(m^2+mn+n^2)}{(m+n)(m-n)}$

Note $\displaystyle m^2-n^2=(m+n)(m-n)$

That leaves us with
$\displaystyle \frac{(m+n)(m-n)(m+n)-(m-n)(m^2+mn+n^2)}{(m+n)(m-n)}$

$\displaystyle \frac{(m+n)^2-(m^2+mn+n^2)}{(m+n)}$

3. $\displaystyle \frac{m^2-n^2}{m-n} - \frac{m^3-n^3}{m^2-n^2}$

common denominator ...

$\displaystyle \frac{(m+n)(m^2-n^2)}{(m+n)(m-n)} - \frac{m^3-n^3}{m^2-n^2}$

$\displaystyle \frac{(m+n)(m^2-n^2)-(m^3-n^3)}{(m+n)(m-n)}$

factor the difference of squares and cubes ...

$\displaystyle \frac{(m+n)(m+n)(m-n)-(m-n)(m^2+mn+n^2)}{(m+n)(m-n)}$

divide out the common factor (m-n) ...

$\displaystyle \frac{(m+n)(m+n)-(m^2+mn+n^2)}{m+n}$

expand (m+n)(m+n) ...

$\displaystyle \frac{(m^2+2mn+n^2)-(m^2+mn+n^2)}{m+n}$

combine like terms ...

$\displaystyle \frac{mn}{m+n}$

4. Thank you, I understand that exercise now
But, I came over this new one:

What is the common denominator here? I could use $\displaystyle {(x+2)(x-2)$ as a common denominator on first two blocks, but the last one has $\displaystyle (x-2)^2}$ under it, how do I deal with that?

This is where I am stuck..

5. Hello, Wraith!

$\displaystyle \text{Simplify: }\;\dfrac{(m+n)(m^2-n^2) - (m-n)(m^2+mn + n^2)}{(m+n)(m-n)}$

We have: .$\displaystyle \dfrac{(m^3 - mn^2 + m^2n - n^3) - (m^3 - n^3)}{(m+n)(m-n)}$

. . . . . . $\displaystyle =\;\dfrac{m^3 - mn^2 + m^2n - n^3 - m^3 + n^3}{(m+n)(m-n)}$

. . . . . . $\displaystyle =\;\dfrac{m^2n - mn^2}{(m+n)(m-n)}$

. . . . . . $\displaystyle =\;\dfrac{mn(m-n)}{(m+n)(m-n)}$

. . . . . . $\displaystyle =\;\dfrac{mn}{m+n}$

6. Originally Posted by Wraith
Thank you, I understand that exercise now
But, I came over this new one:

What is the common denominator here? I could use $\displaystyle {(x+2)(x-2)$ as a common denominator on first two blocks, but the last one has $\displaystyle (x-2)^2}$ under it, how do I deal with that?

This is where I am stuck..
note that the common denominator is $\displaystyle (x+2)(x-2)^2$