Exponent n as a whole number

**p.numminen** · September 29th 2007, 07:38 AM

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.

**earboth** · September 29th 2007, 07:58 AM

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:

$\frac{-a^{2n}}{\frac{1}{a^n}} = -a^{2n+n} = -a^{3n}$

**p.numminen** · September 29th 2007, 08:48 AM

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 $(-a^2)^n \cdot b^{-n}$ ?

**topsquark** · September 30th 2007, 07:18 AM

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.

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

**Soroban** · September 30th 2007, 11:47 AM

Hello, p.numminen!

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

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

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

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