More simplifying help.

**el123** · September 16th 2009, 10:32 PM

Can someone show me how

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

is te he same as

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

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

**Grandad** · September 17th 2009, 12:01 AM

Hello el123

Originally Posted by el123

Can someone show me how

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

is te he same as

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

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:

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

and that's all.

Grandad

**el123** · September 17th 2009, 12:20 AM

that $3x^3$
is supposed to be

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

check it right here.

- Wolfram|Alpha

**Grandad** · September 17th 2009, 12:44 AM

Hello el123

Originally Posted by el123

that $3x^3$
is supposed to be

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

check it right here.

- Wolfram|Alpha

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

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

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

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

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

Grandad

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