Solution of Lagrangian Could you give me advice how to solve the following Lagrangian?
$$L=x^3+y^3 - \lambda (x^2-xy+y^2-5)$$
$$ \left\{ \begin{array}{c} 
\frac{\partial L}{\partial x} = 3x^2 - \lambda (2x-y) = 0
\\\frac{\partial L}{\partial y} = 3y^2 - \lambda (-x+2y) = 0
\\ \frac{\partial L}{\partial \lambda} = x^2-xy+y^2=5
\end{array} \right. $$
$$ \left\{ \begin{array}{c} 
3x^2 / (2x-y)= \lambda 
\\ 3y^2 / (-x+2y) = \lambda 
\\  x^2-xy+y^2=5 \Rightarrow x^2 = xy-y^2+5
\end{array} \right. $$
$$\Rightarrow \lambda  = 3x^2 / (2x-y)= 3y^2 / (-x+2y) \Rightarrow x^2 = \frac{(2x-y)(y^2)}{2y-x} \Rightarrow \frac{(2x-y)(y^2)}{2y-x} = xy-y^2+5 \Rightarrow (2x-y)(y^2) = (xy-y^2+5) (2y-x) \Rightarrow 2xy^2 - y^3 = 2xy^2 - 2y^3 +10y - x^2y +xy^2 -5x \Rightarrow  0 =  - y^3 +10y - x^2y +xy^2 -5x $$
I don't know how to transform the above in order to find x,y.

First
$$\lambda = \frac{3x^2}{2x-y} = \frac{3y^2}{-x+2y}\implies -x^3 +2x^2y = 2y^2x - y^3.$$
Using the factoring formula for $x^n -y^n$ when $n$ is an odd number:
$$
x^n - y^n = (x-y)(x^{n-1} +  x^{n-2}y + \cdots + xy^{n-2} + y^{n-1}).
$$
Hence we have
$$
x^3 -y^3 + 2y^2x - 2x^2y = 0\implies (x-y)(x^2-xy+y^2) = 0.
$$
I believe you can take it from here.
