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Math Help - Indices

  1. #1
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    Indices

    Can someone please help me with this question.

    A) if x=2^{1/3} + 4^{1/3} show that x^3 = 6(1+x)

    B) If x= \frac {1}{2} + \frac {1}{2} \sqrt{5}, show that \frac {x^2+x^{-2}}{x-x^{-1}} = 3

    Thanks in advance
    Last edited by deltaxray; November 7th 2009 at 03:13 AM.
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  2. #2
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    Quote Originally Posted by deltaxray View Post
    Can someone please help me with this question.
    A) if x=2^{1/3} + 4^{1/3} show that x^3 = 6(1+x)
    B) If x= \frac {1}{2} + \frac {1}{2} \sqrt{5}, show that \frac {x^2+x^{-2}}{x-x^{-1}} = 3
    For part B) I would note that
    \frac {x^2+x^{-2}}{x-x^{-1}} = \frac {x^4+1}{x^3-x}
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  3. #3
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    Quote Originally Posted by deltaxray View Post
    Can someone please help me with this question.

    A) if x=2^{1/3} + 4^{1/3} show that x^3 = 6(1+x)

    B) If x= \frac {1}{2} + \frac {1}{2} \sqrt{5}, show that \frac {x^2+x^{-2}}{x-x^{-1}} = 3

    Thanks in advance
    x^3 = \left(2^{\frac{1}{3}} + 4^{\frac{1}{3}}\right)^3

     = \left(2^{\frac{1}{3}}\right)^3 + 3\left(2^{\frac{1}{3}}\right)^2\left(4^{\frac{1}{3  }}\right) + 3\left(2^{\frac{1}{3}}\right)\left(4^{\frac{1}{3}}  \right)^2 + \left(4^{\frac{1}{3}}\right)^3

     = 2 + 3\cdot 2^{\frac{2}{3}} \cdot 4^{\frac{1}{3}} + 3 \cdot 2^{\frac{1}{3}}\cdot 4^{\frac{2}{3}} + 4

     = 6 + 3\cdot 2^{\frac{4}{3}} + 3\cdot 2^{\frac{5}{3}}



    6(1 + x) = 6\left(1 + 2^{\frac{1}{3}} + 4^{\frac{1}{3}}\right)

     = 6 + 6\cdot 2^{\frac{1}{3}} + 6\cdot 2^{\frac{2}{3}}

     = 6 + 3\cdot 2 \cdot 2^{\frac{1}{3}} + 3 \cdot 2 \cdot 2^{\frac{2}{3}}

     = 6 + 3\cdot 2^{\frac{4}{3}} + 3\cdot 2^{\frac{5}{3}}

     = x^3.
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