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Math Help - solving for x.....

  1. #1
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    solving for x.....

    I know this sounds a little stupid and it may sound like I am just trying to get answers but really clueless. The only thing I could really think of is factoring out (d-3)^3 and then finding a lowest common denominator so I can multiply by the bottom and get rid of the fraction and then factor out the x's and solve for them... But I keep getting lost in all my work... Would appreciate any help. Thanks!

    Solve for x.

    <br />
0=\frac{-2c}{x^3}+\frac{6c}{(d-x)^3}<br />
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  2. #2
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    Quote Originally Posted by pakman134 View Post
    I know this sounds a little stupid and it may sound like I am just trying to get answers but really clueless. The only thing I could really think of is factoring out (d-3)^3 and then finding a lowest common denominator so I can multiply by the bottom and get rid of the fraction and then factor out the x's and solve for them... But I keep getting lost in all my work... Would appreciate any help. Thanks!

    Solve for x.

    <br />
0=\frac{-2c}{x^3}+\frac{6c}{(d-x)^3}<br />
    0 = -\frac{2c}{x^3} + \frac{6c}{(d - x)^3}

    Multiply both sides by x^3(d - x)^3:
    0 = -2c(d - x)^3 + 6cx^3

    Now expand:
    0 = -2cd^3 + 6cd^2x - 6cdx^2 + 2cx^3 + 6cx^3

    8cx^3 - 6cdx^2 + 6cd^2x - 2cd^3 = 0

    4x^3 - 3dx^2 + 3d^2x - d^3 = 0

    I see no way to continue from here without a MAJOR operation involving something like Cardano's method.

    -Dan
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  3. #3
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    First combine the fractions

    \frac{6c}{(d-x)^3}- \frac{2c}{x^3}  \Rightarrow \frac{6cx^3 - 2c(d-x)^3}{(d-x)^3x^3}

    so we have \frac{6cx^3 - 2c(d-x)^3}{(d-x)^3x^3} = 0

    therefore 6cx^3 - 2c(d-x)^3 = 0
    \Rightarrow  3x^3 - (d-x)^3 = 0
    \Rightarrow  (3^{\frac{1}{3}}x)^3 - (d-x)^3 = 0

    now you use the difference of two cubes

    (3^{\frac{1}{3}}x -(d-x))((3^{\frac{1}{3}}x)^2 + 3^{\frac{1}{3}}x(d-x) + (d-x)^2 ) = 0

    deal with the linear factor to get x = \frac{d}{3^{\frac{1}{3}} + 1}
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