Identity proof

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

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


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

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

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

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

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

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

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

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

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

$\displaystyle x=1$


So identity is not valid.
Am I right?

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?

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Originally Posted by ThePerfectHacker

Did you make a mistake when you multiplied the 3 terms by the 4 terms?

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I don't think so, I checked.

Quote:

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Originally Posted by ThePerfectHacker

And I also do not understand what all that stuff in the beginning is?

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What do you mean exactly?

I also checked, confirmed.

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

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Originally Posted by DenMac21

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

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


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

[...skip...]

$\displaystyle x=1$


So identity is not valid.
Am I right?

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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.

Quote:

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Originally Posted by ThePerfectHacker

I also checked, confirmed.

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

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The "$\displaystyle |x| <> 1, y <> 0, y<>-x$" is meant to be the set of conditions:

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

RonL