1. ## Identity proof

I have to prove that if $|x| <> 1, y <> 0, y<>-x$ then identity is valid:

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

I have try to prove it.This what I have done:

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

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

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

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

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

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

$-1+xy^3+y^4-y=(y^2-y+xy-x)(y^2+y+1)$

$-1+xy^3+y^4-y=-x+xy^3+y^4-y$

$x=1$

So identity is not valid.
Am I right?

2. Did you make a mistake when you multiplied the 3 terms by the 4 terms?
And I also do not understand what all that stuff in the beginning is?

3. Originally Posted by ThePerfectHacker
Did you make a mistake when you multiplied the 3 terms by the 4 terms?
I don't think so, I checked.

Originally Posted by ThePerfectHacker
And I also do not understand what all that stuff in the beginning is?
What do you mean exactly?

4. I also checked, confirmed.

That means that if the identity is true then $x=1$ which means that if $x$ is anything else from 1 then it is false. Which means it cannot be an identity.

5. Originally Posted by DenMac21
I have to prove that if $|x| <> 1, y <> 0, y<>-x$ then identity is valid:

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

I have try to prove it.This what I have done:

[...skip...]

$x=1$

So identity is not valid.
Am I right?
As in proving trig identities, here it is better to check first if the probvlem identity is true for a numerical substitution.
Try x=0 and y=2.
You get the LHS to be 13/4, while the RHS to be 7/2.
LHS <> RHS, hence, not an identity.

So no need to crack cranium for those x's and y's.

6. Originally Posted by ThePerfectHacker
I also checked, confirmed.

That means that if the identity is true then $x=1$ which means that if $x$ is anything else from 1 then it is false. Which means it cannot be an identity.
The " $|x| <> 1, y <> 0, y<>-x$" is meant to be the set of conditions:

$x^2 \ne 1, y \ne 0, y \ne -x$

RonL