If $u$, $v$ and $w$ are the roots of the cubic $(\lambda - x)^3 + (\lambda - y)^3 + (\lambda - z)^3 = 0$ in $\lambda$, then find partial derivative of $u$, $v$ and $w$ with respect to $x$, $y$ and $z$.
More importantly, I'd like to know how could I express $u$, $v$ and $w$ in terms of $x$, $y$ and $z$?
My thoughts (ie, a possible hint)
Expanding the above, I found (please check) $$3 \lambda^3 - (x+y+z)\lambda^2-[(x^2-2x)+(y^2-2y)+(z^2-2z)] \lambda + 2(x^2+y^2+z^2)-(x^3+y^3+z^3)=0$$
Write \begin{align} p(x, y, z) &= (x+y+z)\\ q(x,y , z) &= -(x^2-2x)-(y^2-2y)-(z^2-2z) \\ r(x, y, z) &= 2(x^2+y^2+z^2)-(x^3+y^3+z^3) \end{align} Now, by Vieta's formula (and using your notation for roots) \begin{align} u+v+w &= (x+y+z)\\ uv+vw+wu &= -(x^2-2x)-(y^2-2y)-(z^2-2z) \\ uvw &= 2(x^2+y^2+z^2)-(x^3+y^3+z^3) \end{align} Now it is tedious and fellow posters may point you toward a more elegant way of finding $u, v, w$ but this way is a useful exercise in algebra!