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Thread: a^4+1/a^4=119. Then a^3+1/a^3

  1. #1
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    a^4+1/a^4=119. Then a^3+1/a^3

    $a^4+\frac{1}{a^4}=119$Then find $a^3+\frac{1}{a^3}$

    I know $a^3+b^3=(a+b)(a^2+b^2-ab)$ and $a^4+b^4+a^2b^2=(a^2+b^2+ab)(a^2+b^2-ab)$

    I try combining both but end up with equation in terms of a to the power 8, 7, 6 etc.
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  2. #2
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    Re: a^4+1/a^4=119. Then a^3+1/a^3

    $\displaystyle a^4+\dfrac{1}{a^4}+1 = \left( a^2+\dfrac{1}{a^2}+1 \right)\left( a^2+\dfrac{1}{a^2} - 1 \right) = 119+1 = 120$

    Let $\displaystyle x = a^2+\dfrac{1}{a^2}$. Then $\displaystyle (x+1)(x-1) = 120$. So, $\displaystyle x^2-1 = 120$ and $\displaystyle x^2 = 121$. This gives $\displaystyle x = \pm 11$.

    Then $\displaystyle \pm 11 = a^2+\dfrac{1}{a^2}$, so $\displaystyle a^4\pm 11a^2+1=0$. Then $\displaystyle a^2 = \dfrac{11 \pm \sqrt{117} }{2}$ or $\displaystyle a^2 = \dfrac{ -11 \pm \sqrt{117} }{2}$. Since any real number squared must be positive, it must be the first one.

    So, $\displaystyle a = \sqrt{ \dfrac{ 11\pm \sqrt{117} }{2} }$ or $\displaystyle a = -\sqrt{ \dfrac{ 11\pm \sqrt{117} }{2} }$. This gives:

    $\displaystyle \begin{align*}\left|a^3+\dfrac{1}{a^3}\right| & = \left( \dfrac{ 11\pm \sqrt{117} }{2} \right)^{3/2} + \left( \dfrac{ 11\pm \sqrt{117} }{2} \right)^{-3/2} \\ & = \left( \dfrac{ 11\pm 3\sqrt{13} }{2}\cdot \dfrac{2}{2} \right)^{3/2} + \left( \dfrac{2}{ 11\pm 3\sqrt{13} }\cdot \dfrac{ 11\mp 3\sqrt{13} }{ 11\mp 3\sqrt{13} } \right)^{3/2} \\ & = \left( \dfrac{ 22\pm 6\sqrt{13} }{4} \right)^{3/2} + \left( \dfrac{ 22 \mp 6\sqrt{13} }{ 4 } \right)^{3/2}\end{align*}$

    $\displaystyle 22\pm 6\sqrt{13} = 9\pm 6\sqrt{13} + 13 = (\sqrt{13} \pm 3)^2$

    Since the square root of a positive number is positive, I put the $\displaystyle \sqrt{13}$ first since $\displaystyle \sqrt{13}-3>0$ and $\displaystyle (\sqrt{13} \pm 3)^2 = (3\pm \sqrt{13})^2$.

    Hence, we have

    $\displaystyle \begin{align*}\left( \dfrac{ 22\pm 6\sqrt{13} }{4} \right)^{3/2} + \left( \dfrac{ 22 \mp 6\sqrt{13} }{ 4 } \right)^{3/2} & = \dfrac{ \left[\left(\sqrt{13}\pm 3\right)^2\right]^{3/2} + \left[\left(\sqrt{13}\mp 3\right)^2\right]^{3/2} }{2^3} \\ & = \dfrac{ \left(13\sqrt{13} \pm 3\cdot 3\cdot 13 + 3\cdot 3^2\sqrt{13} \pm 3^3 \right) + \left(13\sqrt{13} \mp 3\cdot 3\cdot 13 + 3\cdot 3^2\sqrt{13} \mp 3^3\right) }{8} \\ & = \dfrac{80\sqrt{13}}{8} \\ & = 10\sqrt{13}\end{align*}$

    Hence, $\displaystyle a^3+\dfrac{1}{a^3} = \pm 10\sqrt{13}$.
    Thanks from NameIsHidden and topsquark
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    Re: a^4+1/a^4=119. Then a^3+1/a^3

    Quote Originally Posted by SlipEternal View Post
    $\displaystyle a^4+\dfrac{1}{a^4}+1 = \left( a^2+\dfrac{1}{a^2}+1 \right)\left( a^2+\dfrac{1}{a^2} - 1 \right) = 119+1 = 120$

    Let $\displaystyle x = a^2+\dfrac{1}{a^2}$. Then $\displaystyle (x+1)(x-1) = 120$. So, $\displaystyle x^2-1 = 120$ and $\displaystyle x^2 = 121$. This gives $\displaystyle x = \pm 11$.

    Then $\displaystyle \pm 11 = a^2+\dfrac{1}{a^2}$, so $\displaystyle a^4\pm 11a^2+1=0$. Then $\displaystyle a^2 = \dfrac{11 \pm \sqrt{117} }{2}$ or $\displaystyle a^2 = \dfrac{ -11 \pm \sqrt{117} }{2}$. Since any real number squared must be positive, it must be the first one.

    So, $\displaystyle a = \sqrt{ \dfrac{ 11\pm \sqrt{117} }{2} }$ or $\displaystyle a = -\sqrt{ \dfrac{ 11\pm \sqrt{117} }{2} }$. This gives:

    $\displaystyle \begin{align*}\left|a^3+\dfrac{1}{a^3}\right| & = \left( \dfrac{ 11\pm \sqrt{117} }{2} \right)^{3/2} + \left( \dfrac{ 11\pm \sqrt{117} }{2} \right)^{-3/2} \\ & = \left( \dfrac{ 11\pm 3\sqrt{13} }{2}\cdot \dfrac{2}{2} \right)^{3/2} + \left( \dfrac{2}{ 11\pm 3\sqrt{13} }\cdot \dfrac{ 11\mp 3\sqrt{13} }{ 11\mp 3\sqrt{13} } \right)^{3/2} \\ & = \left( \dfrac{ 22\pm 6\sqrt{13} }{4} \right)^{3/2} + \left( \dfrac{ 22 \mp 6\sqrt{13} }{ 4 } \right)^{3/2}\end{align*}$

    $\displaystyle 22\pm 6\sqrt{13} = 9\pm 6\sqrt{13} + 13 = (\sqrt{13} \pm 3)^2$

    Since the square root of a positive number is positive, I put the $\displaystyle \sqrt{13}$ first since $\displaystyle \sqrt{13}-3>0$ and $\displaystyle (\sqrt{13} \pm 3)^2 = (3\pm \sqrt{13})^2$.

    Hence, we have

    $\displaystyle \begin{align*}\left( \dfrac{ 22\pm 6\sqrt{13} }{4} \right)^{3/2} + \left( \dfrac{ 22 \mp 6\sqrt{13} }{ 4 } \right)^{3/2} & = \dfrac{ \left[\left(\sqrt{13}\pm 3\right)^2\right]^{3/2} + \left[\left(\sqrt{13}\mp 3\right)^2\right]^{3/2} }{2^3} \\ & = \dfrac{ \left(13\sqrt{13} \pm 3\cdot 3\cdot 13 + 3\cdot 3^2\sqrt{13} \pm 3^3 \right) + \left(13\sqrt{13} \mp 3\cdot 3\cdot 13 + 3\cdot 3^2\sqrt{13} \mp 3^3\right) }{8} \\ & = \dfrac{80\sqrt{13}}{8} \\ & = 10\sqrt{13}\end{align*}$

    Hence, $\displaystyle a^3+\dfrac{1}{a^3} = \pm 10\sqrt{13}$.
    THanks and very very sorry. I asked my friend in phone. He said that we should find a^3-1/a^3 instead of a^3+1/a^3.
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  4. #4
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    Re: a^4+1/a^4=119. Then a^3+1/a^3

    $\displaystyle a^3-b^3 = (a-b)(a^2+b^2+ab)$

    So, we have

    $\displaystyle \begin{align*}\left| a-\dfrac{1}{a} \right| & = \sqrt{ \dfrac{ 11\pm 3\sqrt{13} }{2} } - \sqrt{ \dfrac{2}{ 11 \pm 3\sqrt{13} } } \\ & = \dfrac{ \sqrt{13} \pm 3 }{2} - \dfrac{ \sqrt{13} \mp 3 }{2} \\ & = \pm 3\end{align*}$

    Above, I showed $\displaystyle x = a^2+\dfrac{1}{a^2} = 11$.

    So, $\displaystyle a-\dfrac{1}{a} = \pm 3$, and $\displaystyle a^3 - \dfrac{1}{a^3} = \left( a - \dfrac{1}{a} \right) \left( a^2 + \dfrac{1}{a^2} + 1\right) = (\pm 3)(11+1) = \pm 36$
    Last edited by SlipEternal; Jul 9th 2014 at 07:46 AM.
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    Re: a^4+1/a^4=119. Then a^3+1/a^3

    Thank you very much.


    I don't understand how the first thing is same as the second line?

    Quote Originally Posted by SlipEternal View Post
    $\displaystyle \begin{align*} \sqrt{ \dfrac{ 11\pm 3\sqrt{13} }{2} } - \sqrt{ \dfrac{2}{ 11 \pm 3\sqrt{13} } } & = \dfrac{ \sqrt{13} \pm 3 }{2} - \dfrac{ \sqrt{13} \mp 3 }{2}\end{align*}$
    [/tex]
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  6. #6
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    Re: a^4+1/a^4=119. Then a^3+1/a^3

    I did all of the work for that in the post above.
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    Re: a^4+1/a^4=119. Then a^3+1/a^3

    Thank you. I did not go through your first post properly.
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  8. #8
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    Re: a^4+1/a^4=119. Then a^3+1/a^3

    Here's a way without actually solving for $\displaystyle a$. First consider

    $\displaystyle \left(a^2 + \dfrac{1}{a^2} \right)^2 = a^4 + 2 + \dfrac{1}{a^4} = 119 + 2 = 121$

    so $\displaystyle a^2 + \dfrac{1}{a^2}=11$ (already established)!

    Then consider

    $\displaystyle \left(a + \dfrac{1}{a} \right)^4 = a^4 + 4a^2 + 6 + \dfrac{4}{a^2} + \dfrac{1}{a^4} = 119 + 4 \cdot 11 + 6 = 169$

    so

    $\displaystyle a + \dfrac{1}{a} = \pm \sqrt{13}$

    Then consider

    $\displaystyle \left(a + \dfrac{1}{a} \right)^3 = a^3 + 3a^2 + \dfrac{3}{a} + \dfrac{1}{a^3} $

    from which we obtain

    $\displaystyle a^3 + \dfrac{1}{a^3} = \left(a + \dfrac{1}{a} \right)^3 - 3a^2 - \dfrac{3}{a} $.

    Then substitute what we know giving

    $\displaystyle a^3 + \dfrac{1}{a^3} = \pm 13\sqrt{13} - 3(\pm\sqrt{13}) = \pm 10\sqrt{13} $.
    Thanks from topsquark and NameIsHidden
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  9. #9
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    Re: a^4+1/a^4=119. Then a^3+1/a^3

    There's a typo that I can't fix. The $\displaystyle 3a^2$ should be a $\displaystyle 3a$.
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