Given z^4 + 2x^2 z^2 + 2y^2 z^2 - 8z = 0, express z as a function of x and y.

Step-by-step solution

1. 1

   Factor out the common $z$. Every term contains at least one $z$:

   $z\left(z^{3}+2x^{2}z+2y^{2}z-8\right)=0\ \Longrightarrow\ z\left(z^{3}+2\left(x^{2}+y^{2}\right)z-8\right)=0$

   so $z=0$ is one branch, valid for every $(x,y)$, and the rest of the work concerns the cubic factor.

2. 2

   Introduce polar radius to expose the structure. The variables enter only through $r^{2}=x^{2}+y^{2}$, so the surface is rotationally symmetric about the $z$-axis and the cubic is

   $z^{3}+2r^{2}z-8=0$

   This is already a depressed cubic $z^{3}+pz+q=0$ (no $z^{2}$ term) with $p=2r^{2}$ and $q=-8$, which is exactly the form Cardano formula handles.

3. 3

   Check how many real roots there are. The cubic discriminant condition uses

   $\left(\frac{q}{2}\right)^{2}+\left(\frac{p}{3}\right)^{3}=16+\frac{8r^{6}}{27}$

   which is positive for every $r$, since both terms are non-negative and the first is $16$. A positive value here means exactly one real root and two complex ones, so the nonzero branch is single-valued — no case analysis needed.

4. 4

   Apply Cardano formula. With $-\dfrac{q}{2}=4$,

   $z=\sqrt[3]{4+\sqrt{16+\frac{8r^{6}}{27}}}+\sqrt[3]{4-\sqrt{16+\frac{8r^{6}}{27}}}$

   The second cube root takes a negative argument (since the square root exceeds 4), which is fine for real cube roots.

5. 5

   Write the answer in the original variables.

   $f(x,y)=\sqrt[3]{4+\sqrt{16+\frac{8\left(x^{2}+y^{2}\right)^{3}}{27}}}+\sqrt[3]{4-\sqrt{16+\frac{8\left(x^{2}+y^{2}\right)^{3}}{27}}}$

6. 6

   Sanity-check at the origin. At $x=y=0$ the cubic is $z^{3}=8$, giving $z=2$; the formula gives $\sqrt[3]{4+4}+\sqrt[3]{4-4}=2+0=2$ ✓. As $r\to\infty$ the two cube roots nearly cancel and $z\to 0^{+}$, matching the cubic $2r^{2}z\approx 8$, i.e. $z\approx 4/r^{2}$ ✓.