# Exponent n as a whole number

• Sep 29th 2007, 08:38 AM
p.numminen
Exponent n as a whole number
How can I simplify $(-a^2 \cdot b^{-1})^n : (ab)^{-n}, n \in \mathbb{Z}, a \neq 0, b \neq 0$ into $-a^{3n}$ ? Please give all stages.
• Sep 29th 2007, 08:58 AM
earboth
Quote:

Originally Posted by p.numminen
How can I simplify $(-a^2 \cdot b^{-1})^n : (ab)^{-n}, n \in \mathbb{Z}, a \neq 0, b \neq 0$ into $-a^{3n}$ ? Please give all stages.

hello,

$(-a^2 \cdot b^{-1})^n : (ab)^{-n} = \frac{-\frac{a^{2n}}{b^n}}{\frac{1}{a^n \cdot b^n}}$. Now cancel ike terms:

$\frac{-a^{2n}}{\frac{1}{a^n}} = -a^{2n+n} = -a^{3n}$
• Sep 29th 2007, 09:48 AM
p.numminen
What rule is there to justify taking the minus sign out of the parentheses?

Shouldn't $(-a^2 \cdot b^{-1})^n$ be rewritten either as $(-1)^n \cdot a^{2n} \cdot b^{-n}$ or as $(-a^2)^n \cdot b^{-n}$ ?
• Sep 30th 2007, 08:18 AM
topsquark
Quote:

Originally Posted by p.numminen
How can I simplify $(-a^2 \cdot b^{-1})^n : (ab)^{-n}, n \in \mathbb{Z}, a \neq 0, b \neq 0$ into $-a^{3n}$ ? Please give all stages.

Quote:

Originally Posted by p.numminen
What rule is there to justify taking the minus sign out of the parentheses?

Shouldn't $(-a^2 \cdot b^{-1})^n$ be rewritten either as $(-1)^n \cdot a^{2n} \cdot b^{-n}$ or as $(-a^2)^n \cdot b^{-n}$ ?

I agree with you. The answer should be:
$(-a^2 \cdot b^{-1})^n : (ab)^{-n} = \frac{(-1)^n \frac{a^{2n}}{b^n}}{\frac{1}{a^n \cdot b^n}} = (-1)^n a^{3n}$

It will only be $-a^{3n}$ when n is odd.

-Dan
• Sep 30th 2007, 12:47 PM
Soroban
Hello, p.numminen!

Quote:

Simplify: . $(-a^2 \cdot b^{-1})^n \div (ab)^{-n}$

We have: . $\left(-\frac{a^2}{b}\right)^n\cdot\frac{1}{(ab)^{-n}} \;=\;(-1)^n\frac{(a^2)^n}{b^n}\cdot (ab)^n$

. . $= \;(-1)^n\cdot\frac{a^{2n}}{b^n}\cdot a^nb^n \;=\;(-1)^na^{3n}$