Let $a,b\in \mathbb{Z}$ satisfy $a \equiv b \equiv 4 \pmod {27}$ and $b\equiv a+3 \pmod 7$ Show that, for any prime $p$, there exist $x,y,z\in\mathbb{Z}_p$ which satisfy $(x^3 - y^3 + a)(x^3 - z^3 + b)(y^3 - z^3 + b - a) = 0$.
The case $p=3$ is easy to take care of, so treat $p\neq 3$. Here is my progress.
Via a primitive root modulo $p$ we see that there are exactly $p$ cubic residues modulo $p$ if $p\equiv 2 \pmod 3$ and exactly $\frac{p+2}{3}$ cubic residues (including $0$) if $p\equiv 1 \pmod 3$. Consider the sets $X = \{x^3: x\in \mathbb{F}_p^{*}\}$, $Y = \{y^3 - a: y\in \mathbb{F}_p\}$ and $Z = \{z^3 - b: z\in \mathbb{F}_p\}$ (we stress that $X$ does not include $0$, so its cardinality is at least $\frac{p-1}{3}$!). In case $X\cap Y \neq \emptyset$ we obtain a solution modulo $p$ to $x^3 - y^3 + a = 0$; in case $X\cap Z \neq \emptyset$ we obtain a solution modulo $p$ to $x^3 - z^3 + b = 0$; otherwise $X\cap (Y\cup Z) = \emptyset$ and since $(X\cup Y \cup Z) \subseteq \mathbb{F}_p$ we must have $$ p \geq |X\cup Y \cup Z| = |X| + |Y\cup Z| = |X| + |Y| + |Z| - |Y\cap Z| \geq \frac{2(p+2)+(p-1)}{3} - |Y\cap Z| $$ and consequently, $|Y\cap Z| \geq 1$, i.e. $Y \cap Z \neq \emptyset$. But the latter actually shows that there is a solution to $y^3 - z^3 + b - a = 0$ modulo $p$! Now suppose we are in the first case, i.e. $x^3 - y^3 + a = 0$ is soluble modulo $p$ (the case with $x^3 - z^3 + b = 0$ is dealt with analogously and we shall not write it down) and let $(x_0,y_0) \in \mathbb{Z}^2$, $p\nmid x_0$ be a solution. Consider the polynomial $f(x) = x^3 - y_0^3 + a$ $-$ it satisfies $|f(x_0)|_p \leq p^{-1} < 1 = |3x_0^2|_p^2 = |f'(x_0)|_p^2$ and so by Hensel's lemma we obtain $c\in \mathbb{Z}_p$ with $f(c) = 0$ and we are done.
However, it still might happen that I am in the third case $y_0^3 - z_0^3 + b - a \equiv 0 \pmod p$ but with $y_0 \equiv z_0 \equiv 0 \pmod p$. Then applying Hensel is troubling.... so how to escape from this?
Any help appreciated!