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Thread: Algebraic manipulation

  1. #1
    Senior Member I-Think's Avatar
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    Algebraic manipulation

    Fairly challenging question here. I'll still try to solve it while I get the forum's help

    Deduce that, if  x=a(b^3-c^3),  y=b(c^3-a^3) and z=c(a^3-b^3), then

    \frac{x^3+y^3+z^3}{xyz}=\frac{a^3+b^3+c^3}{abc}

    Thanks.
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  2. #2
    Master Of Puppets
    pickslides's Avatar
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     <br />
\frac{x^3+y^3+z^3}{xyz}=\frac{(a(b^3-c^3))^3+(b(c^3-a^3))^3+(c(a^3-b^3))^3}{a(b^3-c^3)\times b(c^3-a^3)\times c(a^3-b^3)<br />
} = \dots<br />

    Now start expanding
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  3. #3
    Senior Member I-Think's Avatar
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    Brute force and lengthy expansions
    *Shudders*

    I was hoping there would be a neat solution to this, but if straight expansion's the way to go then so be it.

    Btw
    100th post !Yay!
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  4. #4
    Master Of Puppets
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    Maybe not the only way to solve this but seems to be the most obvious.

    Congrats on the 100th post.
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