Silly technical question about polynomials in Lagrange's "résolution algébrique" I decided that I'd go through Lagrange's "Sur la Résolution Algébrique des Équations". 
On page 3 I got somewhat stuck. Here's the link I've been working with: http://gdz.sub.uni-goettingen.de/dms/load/img/?PPN=PPN308900308&DMDID=DMDLOG_0019
On page 208 (you just click the forward button 3 times), after naming $x=y+z$ for an equation of the form $x^3+nx+p=0$, he gets to the equation $$(*): y^3+z^3+p+(3yz+n)(y+z)=0,$$ which I'm fine with. But then he sets both parts equal to zero with little explanation: $$y^3+z^3+p=0$$ $$3zy+n=0.$$
How can he do that? After messing around with cases for a while, I decided that this is justified by the fact that $y$ and $z$ are variables that are not strictly "decided" by a given $x$, so as long as both parts can be zero, he can "force" them to be zero. That is, as long as he knows a relationship between $y$ and $z$ for which the system works out, it doesn't really matter that there are other solutions for $(*)$, but he doesn't explicitly say this, as far as my french tells me, so I just wanted to confirm that I'm not missing anything major here.
 A: This isn't a silly question, Lagrange does have a slightly irritating tendency to gloss over details in this paper - as you'll see in the coming pages, he's also quite carefree when it comes to dividing by quantities which may or may not equal zero.
It'll help here to state explicitly what is being claimed. We have a value $x$ satisfying:
$$x^3+nx+p=0$$
Obviously there exist numbers $y$ and $z$ such that $y+z=x$. Indeed, there are infinitely many. The question is, are there numbers such that not only $y+z=x$, but also $3yz+n=0$?
The answer is yes. The second equation is $yz=-\frac n 3$, so basically we're asking to find the two numbers $y$ and $z$ given their sum and their product. This is always possible:
$$y+z=s$$
$$yz=p$$
This becomes $z=s-y$ and then $y(s-y)=p$, which is a quadratic equation and therefore does have a (possibly complex, but this wouldn't have bothered Lagrange) solution.
Thus, let $y$ and $z$ be those two numbers (the ones verifying the above system of equations, I mean), and plugging these values into the original equation we are left with $y^3+z^3+p=0$. In detail, since we know that $x^3+nx+p=0$, and that $x=y+z$, we can conclude that, after expanding and gathering terms:
$$y^3+z^3+p+(3yz+n)(y+z)=0$$
But we chose $y$ and $z$ such that $3yz+n=0$, leaving $y^3+z^3+p=0$.
