I am supposed to reduce a surface $x^4+y^4+z^4+w^4+12xyzw=0$ to a surface whose equation is a quadratic in $x^2,y^2,z^2,w^2$ by change of coordinates. Is the following a valid reduction: first of all add $2x^2y^2-2x^2y^2$ and $2z^2w^2-2z^2w^2$ to the equation then we get $(x^2+y^2)^2-2x^2y^2+(z^2+w^2)^2-2z^2w^2+12xyzw=0$.
After this send $(x:y:z:w)$ to $(x^2+y^2,xy,z^2+w^2,zw)$ so that we get $x^2-2y^2+z^2-2w^2+12yw=0$.
So here I still have $yw$. I think I should get rid of it but I couldn't see a way.
Let $(X,Y,Z,W)=(x^2+y^2,xy-3zw,z^2+w^2,zw)$; then \begin{align} &(x^2+y^2)^2-2x^2y^2+(z^2+w^2)^2-2z^2w^2+12xyzw \\ &=X^2-2(Y+3W)^2+Z^2-2W^2+12(Y+3W)W \\ &=X^2-2Y^2-12YW-18W^2+Z^2-2W^2+12YW+36W^2 \\ &=X^2-2Y^2+Z^2+16W^2. \end{align}