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!

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- Sep 16th 2009, 10:32 PMel123More 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 AMGrandad
- Sep 17th 2009, 12:20 AMel123
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 AMGrandad
Hello el123Yes, 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):

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

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

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

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

Grandad