# How do I solve (a+b)^(-c)?

1. Sep 24, 2010

### Pietair

Good day,

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

Thanks.

2. Sep 24, 2010

### Pietair

Re: (a+b)^(-c)

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

3. Sep 24, 2010

Re: (a+b)^(-c)

Well, what does $x ^{-1}$ represent?

4. Sep 24, 2010

### Pietair

Re: (a+b)^(-c)

(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.

5. Sep 24, 2010

Re: (a+b)^(-c)

To be straight that I understand: you have

$$r = \frac 1 {a + b \cos{c}}$$

and need to show that this equals

$$r = \left(\frac 1 a\right) \left( \frac 1 {1 + b \cos{c}}\right)$$

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

6. Sep 24, 2010

### Pietair

Re: (a+b)^(-c)

Exactly.

7. Sep 24, 2010

### Anonymous217

Re: (a+b)^(-c)

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)