There exists soluction by radicals for some class of multivariate complex polynomials?

Am I missing anything?
For any Y,Z,W, just take
X = -(Y + Y\*Z + 10\*Y\*W)/10
and this is a solution (and pretty clearly all of them) yes?
Reduce modulo 10 and you find y(1-z)=0 (mod 10). That gives 4 cases for (y,z) modulo 10. Plugging y=10k gives x=k(10w+z-1). So take any positive k, any positive w, any positive z, and this equation gives you x. For example (w,x,y,z)=(1, 10, 10, 1) is a solution. More generally, (w,k(10w+z-1),10k,z) is a solution for any positive w,k,z. I assume the other 4 cases are equally challenging.
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It’s worth pointing out that even without a change of variables you can solve the original equation by noticing that it is homogeneous in x and y.  This means that if (x,y,z,w) is a solution then so is (dx,dy,z,w) for any d.  In particular, if you divide through by the gcd(x,y) you may assume that gcd(x,y) = 1.

But then notice that in the equation 10x + y - yz - 10yw = 0 you have that y divides 10x.  But if gcd(x,y) = 1 then y divides 10 and so y = 1,2,5 or 10.  You can easily solve the original equation by plugging in the various values of y and then solving for z.

For example if y = 5 then

10x + 5 - 5z - 50w = 0 yields z = 2x - 10w + 1.  So you are free to use any x,w and then you have the solutions (dx, 5d, 2x-10w+1, w) where d,x,w are arbitrary positive integers and gcd(5,x) = 1.

You can do this for each value of y to parametrize all solutions to the original equation.

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