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    http://www.physicsandmathstutor.com/pat/solutions-2006/ If someone can tell me how question 7 works they get a cookie
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    (Original post by giadatsr)
    http://www.physicsandmathstutor.com/pat/solutions-2006/ If someone can tell me how question 7 works they get a cookie
    D are perfect squares and Y are perfect cubes.

    We can clearly see that \sqrt[3]{Y} = k\sqrt{D}, so Y = kD^{\frac{3}{2}}.

    The answer follows.
 
 
 
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