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Hi I have to find the stationary points for $$f(x)= x^4+y^4-(x-y)^2.$$ So far i founded the partial derivatives for $x$ and $y$.

Next step is to solve this system to get my critical points: $$ \left\{ \begin{array}{c} 2x^3-x+y=0 \\ 2y^3-y+x=0 \end{array} \right. $$ But I don't really know how can I solve this system without having to substitute $y=x-2x^3$ and then raise at the power of $3$.

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Add the two equations to get $2x^3=-2y^3$. – Git Gud Jun 15 '14 at 15:59
up vote 2 down vote accepted

Adding the two equations gives $$2x^3+2y^3=0\\ \implies x^3+y^3=0$$ See here:

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