Question Statement:- If $u^3+v^3=x+y$ and $u^2+v^2=x^3+y^3$, show that $$\frac{\partial(u,v)}{\partial(x,y)}=\frac{1}{2}\frac{y^2-x^2}{uv(u-v)}$$
My Solution:-
As the relation that are given between $u,v,x \;\& \; y$ do not seemed to me to be separated so as to get the $u$ and $v$ in the terms of $x\;\&\;y$, so I went with the following approach.
We are given with the following relations:-
$$u^3+v^3=x+y\tag{1}$$ $u^2+v^2=x^3+y^3\tag{2}$
On partially differentiating the above given relations we have,
$$3u^2\frac{\partial u}{\partial{x}}+3v^2\frac{\partial{v}}{\partial{x}}=1\tag{3}$$ $$2u\frac{\partial u}{\partial{x}}+2v\frac{\partial{v}}{\partial{x}}=3x^2\tag{4}$$
The equations $(3)$ and $(4)$ can be written as a matrix as follows $$\begin{bmatrix} 3u^2 & 3v^2\\ 2u & 2v \end{bmatrix} \begin{bmatrix} \dfrac{\partial{u}}{\partial{x}}\\ \dfrac{\partial{v}}{\partial{x}} \end{bmatrix} = \begin{bmatrix} 1 \\ 3x^2 \end{bmatrix} \\ \implies \begin{bmatrix} \dfrac{\partial{u}}{\partial{x}}\\ \dfrac{\partial{v}}{\partial{x}} \end{bmatrix}= \dfrac{1}{6uv(u-v)}\begin{bmatrix} 2v & -2u\\ -3v^2 & 3u^2 \end{bmatrix} \begin{bmatrix} 1 \\ 3x^2 \end{bmatrix}\\ \implies \begin{bmatrix} \dfrac{\partial{u}}{\partial{x}}\\ \dfrac{\partial{v}}{\partial{x}} \end{bmatrix}= \dfrac{1}{6uv(u-v)}\begin{bmatrix} 2v-6ux^2\\ 9u^2x^2-3v^2 \end{bmatrix} $$
So, we get
$$\dfrac{\partial{u}}{\partial{x}}=\dfrac{2v-6ux^2}{6uv(u-v)}\\ \&\\ \dfrac{\partial{v}}{\partial{x}}=\dfrac{9u^2x^2-3v^2}{6uv(u-v)}$$
And due to the symmetry in the equations $(1)$ and $(2)$, we get
$$\dfrac{\partial{u}}{\partial{y}}=\dfrac{2v-6uy^2}{6uv(u-v)}\\ \&\\ \dfrac{\partial{v}}{\partial{y}}=\dfrac{9u^2y^2-3v^2}{6uv(u-v)}$$
So, we get the Jacobian determinant as
$$\frac{\partial(u,v)}{\partial(x,y)}= \begin{vmatrix} \dfrac{\partial{u}}{\partial{x}} & \dfrac{\partial{u}}{\partial{y}}\\ \dfrac{\partial{v}}{\partial{x}} & \dfrac{\partial{v}}{\partial{y}} \end{vmatrix}= \begin{vmatrix} \dfrac{2v-6ux^2}{6uv(u-v)} & \dfrac{2v-6uy^2}{6uv(u-v)}\\ \dfrac{9u^2x^2-3v^2}{6uv(u-v)} & \dfrac{9u^2y^2-3v^2}{6uv(u-v)} \end{vmatrix}=\frac{1}{2}\frac{y^2-x^2}{uv(u-v)}$$
My deal with the question:-
This method seemed a bit too long, can you suggest a shorter method. If the method teaches me something new that would be good.