Algebraic manipulation that I don't understand In the solution of an exercise that I was doing there is the following passage:

Knowing that $D+E+F=0$, from the equation $Dab+Eac+Fbc=0$ we can obtain
$$D=c(a-b) \qquad E=b(c-a) \qquad F=a(b-c)$$
Where we assume that $abc(c-b)(b-a)(c-a) \neq 0$ and that $(D,E,F)\neq (0,0,0)$.

What is the reasonement behind this passage?
I don't know if this is relevant, but just to be clear, the equation $Dab+Eac+Fbc=0$ is the equation of a projective conic, so we can multiply by a nonzero scalar without changing the conic.
 A: This is just a system of linear equations. You can solve it by Gaussian elimintation. The only slight problem is to check whether the equations are independent.
If $a=b=0$ (or in fact any two of $a,b,c$ are zero), then the second equation is void, and every triple with $D+E+F=0$ is a solution. However, your formula only finds $D=E=F=0$, which is not the set of all solutions (even up to constant multiple, as you suggested). So in this case, the formula is wrong; perhaps $a=b=0$ was somehow excluded before. Similarly, if $a=b=c$ (not zero), then the second equation is equivalent to the first one. So once again, there should be infinitely many solutions up to constant multiple, but the formula only found $(0,0,0)$.
If exactly one of $a,b,c$ is zero, for example $a=0$ and $b,c\neq 0$, then the second equation simplifies to $F=0$. So in this case, any triple is a solution where $E=-D$ and $F=0$. In this case, the formula is correct, as it finds $D=-bc, E=bc$ (nonzero pair with $E=-D$), which is the only solution up to constant multiple.
Assume that $a,b,c\neq 0$, and not all are equal, say $b\neq c$. The coefficient matrix of the system of equations is
$\begin{pmatrix} 
1& 1& 1\\
ab& ac& bc
\end{pmatrix}$
Subtract $ab$ times the first row from the second:
$\begin{pmatrix} 
1& 1& 1\\
0& a(c-b)& b(c-a)
\end{pmatrix}$
This is clearly a matrix with rank 2, as $a(c-b)\neq 0$. So up to a constant multiple, there is exactly one nonzero solution (the dimension of the space of solutions is $3-2=1$.) Straightforward calculation shows that $D=c(a-b), E=b(c-a), F=a(b-c)$ is indeed a solution, and it is not zero, because $a(b-c)\neq 0$. So this must be the only nonzero solution up to constant multiple.
So the statement is true iff at most one of $a,b,c$ is zero, and not all of $a,b,c$ are equal.
