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How do I make $y$ the subject of the equation $x^3+y^3-3xy=k$?

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    [Roots of cubic function](http://en.wikipedia.org/wiki/Cubic_function#Roots_of_a_cubic_function)2011-11-23
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    You can use the usual [cubic formula](http://mathworld.wolfram.com/CubicFormula.html), but why would you need to solve for $y$? What do you really want to do?2011-11-23
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    @J.M.: I am trying to find the inverse function of the function $f(x,y)=x^3+y^3-3xy$ given that I know $x$.2011-11-23
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    It's not pretty: [Wolfram alpha](http://www.wolframalpha.com/input/?i=x^3%2By^3-3xy%3Dk+solve+for+y)2011-11-23

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