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!
that $\displaystyle 3x^3 $
is supposed to be
$\displaystyle \frac{3x^3}{xy-1}$
check it right here.
- Wolfram|Alpha
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