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Math Help - Exponent n as a whole number

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by p.numminen View Post
    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}
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  3. #3
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    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} ?
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    Quote Originally Posted by p.numminen View Post
    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 View Post
    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
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  5. #5
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    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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