# More simplifying help.

• Sep 16th 2009, 10:32 PM
el123
More simplifying help.
Can someone show me how

$\displaystyle \frac{4x^3-3x^4y}{(1-xy)^2}$

is te he same as

$\displaystyle \frac{x^3}{(1-xy)^2}-3x^3$

They are apparently alternative ways to write each other but i have no idea how!
• Sep 17th 2009, 12:01 AM
Hello el123
Quote:

Originally Posted by el123
Can someone show me how

$\displaystyle \frac{4x^3-3x^4y}{(1-xy)^2}$

is te he same as

$\displaystyle \frac{x^3}{(1-xy)^2}-3x^3$

They are apparently alternative ways to write each other but i have no idea how!

Sorry, but it just isn't.

You can factorise the numerator:

$\displaystyle \frac{4x^3-3x^4y}{(1-xy)^2}=\frac{x^3(4-3xy)}{(1-xy)^2}$

and that's all.

• Sep 17th 2009, 12:20 AM
el123
that $\displaystyle 3x^3$
is supposed to be

$\displaystyle \frac{3x^3}{xy-1}$

check it right here.

- Wolfram|Alpha
• Sep 17th 2009, 12:44 AM
Hello el123
Quote:

Originally Posted by el123
that $\displaystyle 3x^3$
is supposed to be

$\displaystyle \frac{3x^3}{xy-1}$

check it right here.

- Wolfram|Alpha

Yes, it's true you can write $\displaystyle \frac{4x^3-3x^4y}{(1-xy)^2}$ as $\displaystyle \frac{x^3}{(1-xy)^2}-\frac{3x^3}{xy-1}$ if you need to! How does it work? Well, note first that $\displaystyle (1-xy)^2 = (xy-1)^2$, and then multiply top and bottom of the second fraction by (xy-1):

$\displaystyle \frac{x^3}{(1-xy)^2}-\frac{3x^3}{xy-1}$

$\displaystyle =\frac{x^3}{(1-xy)^2}-\frac{3x^3}{(xy-1)}\color{red}\frac{(xy-1)}{(xy-1)}$

$\displaystyle =\frac{x^3-3x^4y+3x^3}{(1-xy)^2}$

$\displaystyle =\frac{4x^3-3x^4y}{(1-xy)^2}$