What are the conditions for solving (a+b)^(-c)?

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SUMMARY

The discussion centers on the mathematical expression (a+b)^(-c) and its implications when c = 1, leading to the equation r = 1/(a + bcos(c)). The participants analyze the equivalence of two forms of the equation: r = 1/(a + bcos(c)) and r = (1/a) * (1/(1 + bcos(c))). It is concluded that these two expressions are not equal under general conditions, prompting the need for specific counterexamples to demonstrate the impossibility of proving their equivalence.

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Pietair
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Good day,

How do I work out (a+b)^(-c)?

Thanks.
 
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In my case, c = 1, so I have got: (a+b)^-1
 


Well, what does x ^{-1} represent?
 


(1/x).

I have got the equation:
r = 1/(a + bcos(c))

This should be equal to:
r = (1/a) * (1/(1+bcos(c)))

I just can't figure out why.
 


To be straight that I understand: you have

<br /> r = \frac 1 {a + b \cos{c}}<br />

and need to show that this equals

<br /> r = \left(\frac 1 a\right) \left( \frac 1 {1 + b \cos{c}}\right)<br />

If your statements are the ones I've written here, they aren't equal.
 


Exactly.
 


Are there some type of conditions? If not, just write down a counterexample, therefore showing it's impossible to prove it (ie. a not = 1, b, c in reals)
 

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