# exponents

• Apr 6th 2013, 11:41 AM
baldysm
exponents
I have the problem in the book of simplify:

(-a3b3)-2/3

I get :

______1_______
(-a3b3)2/3

= _______1______
(-a2b2)

The books answer is the same except that a2b2 is positive. Where did I go wrong?

Thanks
• Apr 6th 2013, 12:01 PM
Plato
Re: exponents
Quote:

Originally Posted by baldysm
I have the problem in the book of simplify:

(-a3b3)-2/3

I get :

______1_______
(-a3b3)2/3

= _______1______
(-a2b2)

The books answer is the same except that a2b2 is positive. Where did I go wrong?

$\displaystyle {\left( { - {a^3}{b^3}} \right)^{\frac{2}{3}}} = {a^2}{b^2}$

i.e. $\displaystyle (-a)^2=a^2$
• Apr 6th 2013, 02:16 PM
HallsofIvy
Re: exponents
Quote:

Originally Posted by baldysm
I have the problem in the book of simplify:

(-a3b3)-2/3

I get :

______1_______
(-a3b3)2/3

= _______1______
(-a2b2)

You didn't square the "-". This should be $\displaystyle \frac{1}{(-a)^2b^2}$ and, of course, $\displaystyle (-a)^2= a^2$

Quote:

The books answer is the same except that a2b2 is positive. Where did I go wrong?

Thanks
• Apr 6th 2013, 07:55 PM
HowDoIMath
Re: exponents
Plato and HallsofIvy have the right idea. Obviously (-a) 2 = a2, but if it doesn't make sense that (-a 3b3)2/3 = a2b2, think of (-a3*b3) as (-1*a 3 *b3).
(-1) 2/3 = (-1)2 = 1, or (-1)2/3 = (1)1/3 = 1