MHB Simplification and manipulation

[mathnap7]

Hi All,

I've been stuck on a problem and after much simplification, it has been reduced to the following:

Is it possible to factor out the (a^2 - b^2) bit. For example, if a = b, the equation would not be valid. Is there any way to factor it out using manipulation?

Best wishes,

'nap

 

[topsquark]

Hi All,

I've been stuck on a problem and after much simplification, it has been reduced to the following:

Is it possible to factor out the (a^2 - b^2) bit. For example, if a = b, the equation would not be valid. Is there any way to factor it out using manipulation?

Best wishes,

'nap

Well [math]a^2 - b^2 = (a + b)(a - b)[/math] but it sounds like you are after something different. Can you post the whole problem?

-Dan

 

[SuperSonic4]

Hi All,

I've been stuck on a problem and after much simplification, it has been reduced to the following:

Is it possible to factor out the (a^2 - b^2) bit. For example, if a = b, the equation would not be valid. Is there any way to factor it out using manipulation?

Best wishes,

'nap

This is a relatively common expression known as the difference of two squares and is well worth committing to memory to save time

$$a^2-b^2 = (a-b)(a+b)$$

 

[mathnap7]

Apologies, I am not too familiar with using the LaTeX plugin.

\frac{a^2b^3}{2(a^2-b^2)}\frac{(cos\phi -1)}{(a^2sin^2\phi +b^2cos^2\phi}

 

[Evgeny.Makarov]

The expression is
\[
\frac{a^2b^3}{2(a^2-b^2)}\cdot\frac{\cos\phi -1}{a^2\sin^2\phi +b^2\cos^2\phi}
\]
Is there a problem in leaving it like this?

Is it possible to factor out the (a^2 - b^2) bit. For example, if a = b, the equation would not be valid. Is there any way to factor it out using manipulation?

The original expression is indeed undefined when $a=b$, and it is not just a removable singularity. Algebraic manipulations will not make it defined when $a=b$.

It may indeed help if you post the whole problem.