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Math Help - simplying indices

  1. #1
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    simplying indices

     (4y^3)^3 \div  2y^3

     4y^9 \div 2y^3

     4 \div 2 = 2    y^9 \div y^3 = y^6

     2y^6

    is this working right? Because my book says the correct answer is  32y^6 and I dont understand were 32 came from ?
    Last edited by Tweety; January 17th 2009 at 08:55 AM.
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  2. #2
    Junior Member ursa's Avatar
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    is this working right? Because my book says the correct answer is and I dont understand were 32 came from ?
    hi
    you forgot to make 4^3=64 also
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Tweety View Post
     (4y^3)^3 \div  2y^3

     4y^9 \div 2y^3

     4 \div 2 = 2    y^9 \div y^3 = y^6

     2y^6

    is this working right? Because my book says the correct answer is  32y^6 and I dont understand were 32 came from ?
    Note that \left(4y^3\right)^3=4^3y^3=\dots
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  4. #4
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    Quote Originally Posted by ursa View Post
    hi
    you forgot to make 4^3=64 also
    Hi thanks for answering but I was just wondering why in this expression  (3x^2)^3 \div x^4 the number 3 is not 3^3 , ?

    as the correct answer for this is  3x^2 so how comes 4 has to be raised to the third power in the expression above?
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  5. #5
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    the answer is wrong as it should be 3^3.
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