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RokerHRO (talk ) 10:11, 6 August 2010 (UTC) [ reply ]
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{\displaystyle x=u(1-u^{2}/3+v^{2})/3,y=v(1-v^{2}/3+u^{2})/3,z=(u^{2}-v^{2})/3}
we can eliminate
u
{\displaystyle u}
v
{\displaystyle v}
64
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128
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64
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5
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702
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18
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144
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{\displaystyle 64z^{9}-128z^{7}+64z^{5}-702x^{2}y^{2}z^{3}-18x^{2}y^{2}z+144(y^{2}z^{6}-x^{2}z^{6})\ }
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162
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27
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9
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48
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{\displaystyle {}+162(y^{4}z^{2}-x^{4}z^{2})+27(y^{6}-x^{6})+9(x^{4}z+y^{4}z)+48(x^{2}z^{3}+y^{2}z^{3})\ }
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432
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81
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240
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135
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{\displaystyle {}-432(x^{2}z^{5}+y^{2}z^{5})+81(x^{4}y^{2}-x^{2}y^{4})+240(y^{2}z^{4}-x^{2}z^{4})-135(x^{4}z^{3}+y^{4}z^{3})=0.\ }
For example, in Wolfram Mathematica or (free, on-line) Wolfram Alpha ,
Eliminate[{x == u*(1 - u^2/3 + v^2)/3, y == v*(1 - v^2/3 + u^2)/3, z == (u^2 - v^2)/3}, {u, v}] produces an equation of the form
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{\displaystyle p(x,y,z)=q(x,y,z)}
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{\displaystyle p(x,y,z)-q(x,y,z)=0}
MathPerson (talk ) 15:37, 21 April 2021 (UTC) [ reply ]
