Got in stuck with this:
If the roots of the equation in $\lambda$ $$(\lambda -x)^3+(\lambda-y)^3+(\lambda-z)^3=0$$ are $u,v,w$ then prove that the Jacobian $$\frac{\partial(u,v,w)}{\partial(x,y,z)}=-2\frac{(y-z)(z-x)(x-y)}{(v-w)(w-u)(u-v)}.$$
I started with the relations between roots and coefficients: \begin{align} &u+v+w=-3(x+y+z)\\ &uv+uw+vw=3(x^2+y^2+z^2)\\ &uvw=-(x^3+y^3+z^3) \end{align}
But from here dont know how to simplify calculation further. How to proceed ?
On
I got this same question in my assignment, and we can try using a property of jacobian here:
By viete formulas we have relations 3 relations on roots, ie $f_1(x,y,z,u ,v,w), f_2(x,y,z,u,v,w), f_3(x,y,z,u,v,w)$ and then we can use the property:
$$\frac{\partial (u, v, w)}{\partial (x,y,z)} = (-1)^{3}\frac{\frac{\partial (f_1, f_2,f_3)}{\partial (x,y,z)}}{\frac{\partial (f_1, f_2, f_3)}{\partial (u,v,w)}}$$
This relation should make calculation easier!
Starting from the original equation, taking the total derivative gives $$ ((\lambda-x)^2+(\lambda-y)^2+(\lambda-z)^2)d\lambda = (\lambda-x)^2 dx + (\lambda-y)^2 dy + (\lambda-z)^2 dz, $$ so this holds for any of $u,v,w$. Looking at the left-hand side, this is $p'(\lambda)/3$ for a cubic polynomial, which we can write in the form $$ p(\lambda) = 3(\lambda-u)(\lambda-v)(\lambda-w) $$ by examining the original form. Written in this form, it becomes obvious that after differentiating, $$ \frac{1}{3}p'(u) = (u-v)(u-w). $$ Now, the Jacobian is given by $ du \wedge dv \wedge dw = \frac{\partial(u,v,w)}{\partial(x,y,z)} dx \wedge dy \wedge dz $, so you can calculate it by wedging $$ (u-v)(u-w) du = (u-x)^2 dx + (u-y)^2 dy + (u-z)^2 dz $$ and its counterparts together. If you're not used to this formalism, you can instead derive the partial derivatives $$ \frac{\partial u}{\partial x} = -\frac{(u-x)^2}{(u-v)(w-u)} $$ and so on from this equation, then find the Jacobian as the determinant of all these. Either way, you end up with $$ -\frac{1}{(u-v)^2(v-w)^2(w-z)^2} \begin{vmatrix} (u-x)^2 & (u-y)^2 & (u-z)^2 \\ (v-x)^2 & (v-y)^2 & (v-z)^2 \\ (w-x)^2 & (w-y)^2 & (w-z)^2 \end{vmatrix}. $$ How to find the last determinant? It is clearly a polynomial of degree 3 in all the variables, and zero when two of $u,v,w$ are equal or two of $x,y,z$ are equal, so it has a factor $$ (u-v)(v-w)(w-u)(x-y)(y-z)(z-x). $$ Moreover, this has degree 3 in all variables, so this must be the only variable element of the polynomial. All we have to do is plug in some numbers that make the determinant nonzero to find the leading coefficient. Putting $u=x=0$, $v=y=1$, $w=z=-1$ gives $$ \begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 4 \\ 1 & 4 & 0 \end{vmatrix} = 8, $$ while the variable factor evaluates to $4$. Hence $$ \begin{vmatrix} (u-x)^2 & (u-y)^2 & (u-z)^2 \\ (v-x)^2 & (v-y)^2 & (v-z)^2 \\ (w-x)^2 & (w-y)^2 & (w-z)^2 \end{vmatrix} = 2(u-v)(v-w)(w-u)(x-y)(y-z)(z-x). $$ Hence the Jacobian is $$ - \frac{2(u-v)(v-w)(w-u)(x-y)(y-z)(z-x)}{(u-v)^2(v-w)^2(w-z)^2} = -2 \frac{(x-y)(y-z)(z-x)}{(u-v)(v-w)(w-z)}. $$ This may not be the simplest possible way, but the actual calculation is fairly minimal.