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Thread: Indices

  1. #1
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    Indices

    Can someone please help me with this question.

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

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

    Thanks in advance
    Last edited by deltaxray; Nov 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 $\displaystyle x=2^{1/3} + 4^{1/3}$ show that $\displaystyle x^3 = 6(1+x)$
    B) If $\displaystyle x= \frac {1}{2} + \frac {1}{2} \sqrt{5}$, show that $\displaystyle \frac {x^2+x^{-2}}{x-x^{-1}} = 3$
    For part B) I would note that
    $\displaystyle \frac {x^2+x^{-2}}{x-x^{-1}} = \frac {x^4+1}{x^3-x}$
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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 $\displaystyle x=2^{1/3} + 4^{1/3}$ show that $\displaystyle x^3 = 6(1+x)$

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

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

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

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

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



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

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

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

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

    $\displaystyle = x^3$.
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