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Algebra > Customizable Word Problem Solvers > Misc > SOLUTION In the equation w^3 x^3 y^3 = z^3, w^3, x^3, y^3 and z^3 are distinct, consecutiveMay 06, 17 · (xy)^3 (yz)3 (zx)^3 = 3(xy)(yz)(zx) That is it no constraints etc It mentions "This can be done by expanding out the brackets, but there is a more elegant solution" Homework Equations The Attempt at a Solution First of all this only seems to hold in special cases as I have substituted random values for x,y and z and they do not agreeApr 24, 13 · If we change the sign $Y^3X^3=qZ^2$ Then the solutions are of the form $X=qa(2p3as)sc^2$ $Y=q(p2as)pc^2$ $Z=q(p^23aps3a^2s^2)c^3$ Another solution of the equation $X^3Y^3=qZ^2$ $p,s$ integers asked us To facilitate the calculations we make the change $a,b,c$ If the ratio is as follows $q=3t^21$ Search Q X3 Y3 Formula Tbm Isch X^3+y^3+z^3=k