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Math Help - Prove this hardest thing

  1. #1
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    Prove this hardest thing

    Prove that--
    {(3+2*5^1/4) (3-2*5^1/4)}^1/4=( 5^1/4 +1) ( 5^1/4 -1)
    If you cannot understand it ,can u teach me how to write it using mathematical signs.
    But for the time being prove it......
    It is simply impossible to prove it
    I have spent my 1 month on it
    but i assure you that this question is correct
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  2. #2
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    Hello, jashansinghal!

    Wow . . . This is a messy one!


    Prove that: . \left(\frac{3+2\!\cdot\!5^{\frac{1}{4}}} {3 - 2\!\cdot\!5^{\frac{1}{4}}}\right)^{\frac{1}{4}} \;=\; \frac{5^{\frac{1}{4}} +1} {5^{\frac{1}{4}} -1}
    This is equivalent to: . \frac{3 + 2\!\cdot\!5^{\frac{1}{4}}} {3 - 2\!\cdot\!5^{\frac{1}{4}}} \;=\;\left(\frac{5^{\frac{1}{4}} + 1} {5^{\frac{1}{4}} - 1}\right)^4 = \;\frac{\left(5^{\frac{1}{4}}+1\right)^4}{\left(5^  {\frac{1}{4}} - 1\right)^4}



    On the right side, the numerator is: . \left(5^{\frac{1}{4}}\right)^4 + 4\left(5^{\frac{1}{4}}\right)^3 + 6\left(5^{\frac{1}{4}}\right)^2 + 4\left(5^{\frac{1}{4}}\right) + 1

    . . . . . = \;\;5 + 4\cdot5^{\frac{3}{4}} + 6\cdot5^{\frac{1}{2}} + 4\cdot5^{\frac{1}{4}} + 1 \;\;=\;\;6 + 6\cdot5^{\frac{1}{2}} + 4\cdot5^{\frac{3}{4}} + 4\cdot5^{\frac{1}{4}}

    Factor: . 6\left(1 + 5^{\frac{1}{2}}\right) + 4\cdot5^{\frac{1}{4}}\left(5^{\frac{1}{2}}+1\right  ) \;=\;\left(6+4\cdot5^{\frac{1}{4}}\right)\left(5^{  \frac{1}{2}} + 1\right) = \;2\left(3+2\cdot5^{\frac{1}{4}}\right)\left(5^{\f  rac{1}{2}}+1\right)



    On the right side, the denominator is: . \left(5^{\frac{1}{4}}\right)^4 - 4\left(5^{\frac{1}{4}}\right)^3 + 6\left(5^{\frac{1}{4}}\right)^2 - 4\left(5^{\frac{1}{4}}\right) + 1

    . . . . . = \;\;5 - 4\cdot5^{\frac{3}{4}} + 6\cdot5^{\frac{1}{2}} - 4\cdot5^{\frac{1}{4}} + 1 \;\;=\;\;6 + 6\cdot5^{\frac{1}{2}} - 4\cdot5^{\frac{3}{4}} - 4\cdot5^{\frac{1}{4}}

    Factor: . 6\left(1 + 5^{\frac{1}{2}}\right) - 4\cdot5^{\frac{1}{4}}\left(5^{\frac{1}{2}}+1\right  ) \;=\;\left(6-4\cdot5^{\frac{1}{4}}\right)\left(5^{\frac{1}{2}} + 1\right) = \;2\left(3-2\cdot5^{\frac{1}{4}}\right)\left(5^{\frac{1}{2}}+  1\right)



    The fraction on the right becomes: . \frac{2\left(3+2\cdot5^{\frac{1}{4}}\right)\left(5  ^{\frac{1}{2}}+1\right)} {2\left(3-2\cdot5^{\frac{1}{4}}\right)\left(5^{\frac{1}{2}}+ 1\right)} = \;\frac{3+2\cdot5^{\frac{1}{4}}}{3-2\cdot5^{\frac{1}{4}}}


    Hence, we have: . \left(\frac{5^{\frac{1}{4}}+1}{5^{\frac{1}{4}}-1}\right)^4 \;=\; \frac{3+2\cdot5^{\frac{1}{4}}}{3-2\cdot5^{\frac{1}{4}}}


    Therefore: . \frac{5^{\frac{1}{4}}+1}{5^{\frac{1}{4}}-1} \;=\;\left(\frac{3+2\cdot5^{\frac{1}{4}}}{3-2\cdot5^{\frac{1}{4}}}\right)^{\frac{1}{4}}

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  3. #3
    Super Member malaygoel's Avatar
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    Let y=\frac{(x+1)^4}{(x-1)^4}

    \frac{y+1}{y-1}=\frac{(x+1)^4+(x-1)^4}{(x+1)^4-(x-1)^4}

    \frac{y+1}{y-1}=\frac{x^4+6x^2+1}{4x^3+4x}

    substitute x=5^{\frac{1}{4}}

    \frac{y+1}{y-1}=\frac{6+6x^2}{4x^3+4x}

    \frac{y+1}{y-1}=\frac{6}{4x}

    \frac{y+1}{y-1}=\frac{3}{2\cdot5^{\frac{1}{4}}}

    y=\frac{3+2\cdot5^{\frac{1}{4}}}{3-2\cdot5^{\frac{1}{4}}}
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