Proving the Equality of Two Complex Expressions Using Algebraic Manipulation

[Granger]

Homework Statement

I have to prove that the expression

$$\frac{\omega C - \frac{1}{\omega L}}{\omega C - \frac{1}{\omega L} + \omega L - \frac{1}{\omega C}}$$

is equal to

$$\frac{1}{3-( (\frac{\omega_r}{\omega})^2 + (\frac{\omega}{\omega_r})^2)}$$

where $\omega_r= \frac{1}{\sqrt{LC}}$

Homework Equations

3. The Attempt at a Solution [/B]

What I started to do was to get rid of the denominators in the fraction and put everything together

$$\frac{\omega^2C^2L-C}{\omega^2C^2L-C+\omega^2CL^2-L}$$

Then I divided the denominator by the numerator

$$\frac{1}{1+\frac{\omega^2CL^2-L}{\omega^2C^2L-C}}$$

And I'm kind of stuck now. Can someone give an hint on how should I proceed next?
Or is there any easier way to start the proof? I'm just looking for a hint, thanks.

 

[Dick]

Homework Statement

I have to prove that the expression

$$\frac{\omega C - \frac{1}{\omega L}}{\omega C - \frac{1}{\omega L} + \omega L - \frac{1}{\omega C}}$$

is equal to

$$\frac{1}{3-( (\frac{\omega_r}{\omega})^2 + (\frac{\omega}{\omega_r})^2)}$$

where $\omega_r= \frac{1}{\sqrt{LC}}$

Homework Equations

3. The Attempt at a Solution [/B]

What I started to do was to get rid of the denominators in the fraction and put everything together

$\frac{\omega^2C^2L-C}{\omega^2C^2L-C+\omega^2CL^2-L}$

Then I divided the denominator by the numerator

$\frac{1}{1+\frac{\omega^2CL^2-L}{\omega^2C^2L-C}}$

And I'm kind of stuck now. Can someone give an hint on how should I proceed next?
Or is there any easier way to start the proof? I'm just looking for a hint, thanks.

You could start by checking for a typo in the expressions you gave. If you put $L=C=1$ you can easily see that they aren't equal.

 

[Orodruin]

First of all, get rid of L and C in favour of $\omega_r$ as soon as possible. It will save you a lot of writing and make your goal clearer. Second, your desired expression has a 3 in the denominator and you have a 1. How can you compensate for this?

I also suggest working with the reciprocal of your expression. It will save you having to write the expression as a quotient.