MHB Simplification and manipulation

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The discussion centers on the possibility of factoring the expression (a^2 - b^2) in a given mathematical context. Participants clarify that (a^2 - b^2) can be factored as (a + b)(a - b), but emphasize that this does not resolve the issue when a equals b, as the expression becomes undefined. The original poster is encouraged to share the complete problem for further assistance. It is noted that algebraic manipulation cannot make the expression valid when a equals b. The conversation highlights the importance of understanding singularities in mathematical expressions.
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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
 
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mathnap7 said:
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
 
mathnap7 said:
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

$$a^2-b^2 = (a-b)(a+b)$$
 
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}
 
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.
 
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