1. ## Simplifying Rationals

$\frac {5x^{3a} + 5x^{2a} y^a}{x^{2a} - y^{2a}} = \frac {5x^{2a}}{x^a-y^a}$ right?

$\frac {m^3 + n^3}{mp-mq-np+nq} \div \frac {mn-m^2-n^2}{mp-mq+np-nq}$ $\Longrightarrow \frac {(m+n)(m^2-mn+n^2)}{(p-q)(m-n)} \cdot \frac{(p-q)(m+n)}{mn-[(m+n)(m-n)]}$ $\Longrightarrow \frac {m^2 -mn + n^2}{m-n} \cdot \frac {m+n}{mn-(m-n)}$ $= \frac {m^3+n^3}{-m^2+m^2n-mn^2+n^2}$ Did I do this right?

2. Originally Posted by chrozer
$\frac {5x^{3a} + 5x^{2a} y^a}{x^{2a} - y^{2a}} = \frac {5x^{2a}}{x^a-y^a}$ right?

$\frac {m^3 + n^3}{mp-mq-np+nq} \div \frac {mn-m^2-n^2}{mp-mq+np-nq}$ $\Longrightarrow \frac {(m+n)(m^2-mn+n^2)}{(p-q)(m-n)} \cdot \frac{(p-q)(m+n)}{mn-[(m+n)(m-n)]}$ $\Longrightarrow \frac {m^2 -mn + n^2}{m-n} \cdot \frac {m+n}{mn-(m-n)}$ $= \frac {m^3+n^3}{-m^2+m^2n-mn^2+n^2}$ Did I do this right?
THe first one is correct you factored the difference of squares on the bottom and factored out the common factor on top and cancelled right?

3. Originally Posted by Mathstud28
THe first one is correct you factored the difference of squares on the bottom and factored out the common factor on top and cancelled right?
Yep. How about the 2nd one?

4. Originally Posted by chrozer
$\frac {5x^{3a} + 5x^{2a} y^a}{x^{2a} - y^{2a}} = \frac {5x^{2a}}{x^a-y^a}$ right?

$\frac {m^3 + n^3}{mp-mq-np+nq} \div \frac {mn-m^2-n^2}{mp-mq+np-nq}$ $\Longrightarrow \frac {(m+n)(m^2-mn+n^2)}{(p-q)(m-n)} \cdot \frac{(p-q)(m+n)}{mn-[(m+n)(m-n)]}$ $\Longrightarrow \frac {m^2 -mn + n^2}{m-n} \cdot \frac {m+n}{mn-(m-n)}$ $= \frac {m^3+n^3}{-m^2+m^2n-mn^2+n^2}$ Did I do this right?
No...I think you did the second one incorrectly you cancelled m+n in the second things denominator but it was a binomial you cannot cancel this way you...I know you know that you just might have missed it.......for example

$\frac{a}{b}\cdot{b}=a$

but $\frac{a}{b+a}\cdot{b}\ne\frac{a}{a}$ you cant cancel when theyre are terms like that

5. Originally Posted by Mathstud28
No...I think you did the second one incorrectly you cancelled m+n in the second things denominator but it was a binomial you cannot cancel this way you...I know you know that you just might have missed it.......for example

$\frac{a}{b}\cdot{b}=a$

but $\frac{a}{b+a}\cdot{b}\ne\frac{a}{a}$ you cant cancel when theyre are terms like that
What else can i cancel out other than the $(p-q)$ ?

6. Originally Posted by chrozer
What else can i cancel out other than the $(p-q)$ ?
Nothing that I can see of

7. Originally Posted by Mathstud28
Nothing that I can see of
So would the answer equal $\frac {(m+n)^2(m^2-mn+n^2)}{(m-n)(mn-m^2-n^2)}$ ?

8. Originally Posted by chrozer
So would the answer equal $\frac {(m+n)^2(m^2-mn+n^2)}{(p-q)(m-n)(mn-m^2-n^2)}$ ?
I believe so...I could be wrong though

9. Originally Posted by Mathstud28
I believe so...I could be wrong though
NVM. I found the right answer. It is $-\frac {(m+n)^2}{(m-n)}$.

10. Originally Posted by chrozer
NVM. I found the right answer. It is $-\frac {(m+n)^2}{(m-n)}$.
Ya that is correct.

Mathstud28: You didnt notice $m^2 -mn + n^2 = -( mn - m^2 - n^2)$ ??