Four variables and one equation?

Let $x,y,z,p$ be a solution of the above diophantine equation. We can write $x$ as $x=\pm p^m r$ , where $m \geq 0$ is the highest power of $p$ that divides $x$ (This can be zero, if $x$ is not divisible by $p$ ). From the given equation, it is easy to see that $p|x^3$ , i.e. $p|x$ . So, let $x=px_1$ , where $x_1$ is an integer. Then we have the following equation

$p^3x_1^3+y^3p+z^3p^2=0$ ,

i.e. $y^3+z^3p+x_1^3p^2=0$

This implies that $p|y$ , So let $y=py_1$ , where $y_1$ is an integer. We have the similar equation again $z^3+x_1^3p+y_1^3p^2=0$ . Thus, $p|z$ . Let $z=pz_1$ , with $z_1$ being an integer. Hence we have,

$x_1^3+y_1^3p+z_1^3p^2=0$

Which has the exact same form of the diphantine equation we started with. Hence we can repeat this step more than $m+1$ times, to conclude that $p^{m+1}|x$ , which contradicts the fact that $m$ is the highest power of $p$ that divides $x$ .

Hence no solution is possible.

X^3 = -p(y^3+z^3 p) So p l x^3 then p l x Same way p l y , p l z X^3 + y^3 p + z^3 p^2 = (Pk)^3 + (pl)^3 p + (pm)^3*p^2

(k)^3 + (l)^3 p + (m)^3 p^2 is same type pf above equation. so, X=p^a y=p^b z=p^c. X^3+ y^3 p +z^3 p^2 = p^{3a}+p^{3b+1}+p^{3c+2} =0 Divide all terms with p^{3b+1}, p^{3(a-b)-1}+ p^{3(c-b)+1} = -1. Above equation is always fault whatever a,b,c is. So there are no solution (x,y,z,p).