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Math Help - Simplification

  1. #1
    Newbie
    Joined
    May 2007
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    Simplification

    Hello,

    If  \frac{a}{b} = \frac{b}{c} = \frac{c}{d}, prove that
     \frac{a}{d} is equal to  \sqrt{\frac{a^5 + b^2c^2 + a^3c^2}{b^4c + d^4 + b^2cd^2}}

    It seems like a mess. Any suggestions on where to start?

    Btw, this problem is from Higher Algebra by H.S Hall and S. R Knight, a textbook that is still used in India even though it was written in the 1800's.
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  2. #2
    Senior Member JaneBennet's Avatar
    Joined
    Dec 2007
    Posts
    293
    We are assuming that a,b,c,d>0, right? (If not, you could have, say, a=c=1 and b=d=-1 and then \frac{a}{d} would be negative.)

    Looking at the expression, the best thing to do first looks like eliminating b^2=ac:

    \color{white}.\quad. \frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}

    =\ \frac{a^5+ac^3+a^3c^2}{a^2c^3+d^4+ac^2d^2}

    We also have a^2d^2=b^2c^2=ac^3, i.e. c^3=ad^2. So

    \color{white}.\quad. \frac{a^5+ac^3+a^3c^2}{a^2c^3+d^4+ac^2d^2}

    =\ \frac{a^5+a^2d^2+a^3c^2}{a^3d^2+d^4+ac^2d^2}

    =\ \frac{a^2\left(a^3+d^2+ac^2\right)}{d^2\left(a^3+d  ^2+ac^2\right)}

    =\ \frac{a^2}{d^2}
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