Prime numbers $(x,c,p)$ such that $x^3-p x^2-cx-5c=0$ How should I proceed to find all prime numbers $x,c,p$ such that
$$x^3-px^2-cx-5c=0$$

If $x^3-px^2-cx-5c=0$, then $x$ divides $5c$. Since $x$ and $c$ are prime, we have the two possibilities $x=c$ and $x=5$. 
Suppose $x=c$. Substitute.  We get $c^3-(p+1)c^2-5c=0$, so $c^2-(p+1)c-5=0$, so $c$ divides $5$, so $c=5$.  Therefore definitely $x=5$. 
Substitute. We get $125-25p-10c=0$. Since $25$ divides $10c$, it follows that $c=5$. Thus $p=3$.  
