Depressed cubic equation, del Farro's calculation I am reading about the solution for depressed cubic at http://fermatslasttheorem.blogspot.ca/2006/11/depressed-cubic.html
One thing I didn't quite understand is at step 3: 
if 
$(3uv + b) = 0$
then:
...
then the solution is derived. 
But, how about the situation for $3uv + b \neq 0$? how do we deal with it?
Thanks for your help.
 A: @Wei Ma, great question! :) Let me explain briefly using theory, and then do a practical example with the quadratic to save on space. :)
As you know, if you have $N$ unknowns, you need at least $N$ equations to solve for those unknowns explicitly. Thus, if you assume: $x=u+v$, you have increased the number of unknowns by two, so you need another equation (since the assumption is one) to solve. We pick the coefficient on the quadratic because $3uv+b=0$ has only one solution if solved for either $u$ or $v$. If we took the coefficients in front of the linear term, there would be a pair of (complex) solutions.
Example: The substitution that you mentioned can also be applied to the quadratic formula for a more direct example. :)
$$ax^2+bx+c=0$$
$$LET : x=u+v$$
$$a(u+v)^2+b(u+v)+c=0$$
$$a(u^2+2uv+v^2)+b(u+v)+c=0$$
$$au^2+2avu+av^2+bu+bv+c=0$$
$$(a)u^2+(2av+b)u+(av^2+bv+c)=0$$
Notice how we have factored for $u$. We need to make an assumed equation to eliminate on of these terms. Clearly, we can't throw out $a$, since we didn't assume it. That only leaves $2av+b=0$ or $2v^2+bv+c=0$. The latter has two solutions, the former has one, so we know we aren't creating problems if we choose it. We are only pushing one value of $v$ to be 0, rather than two. So let's step along.
$$LET : 2av+b=0 \rightarrow v=\frac{-b}{2a}$$
$$(a)u^2+(0)u+(a(\frac{-b}{2a})^2+b\frac{-b}{2a}+c)=0$$
$$(a)u^2+(\frac{ab^2}{4a^2}-\frac{b^2}{2a}+c)=0$$
Notice that we now have the form: $\alpha x^2+\beta=0$. This is trivial to solve, so let's do it and then substitute back.
$$u^2=\frac{b^2}{2a^2}-\frac{b^2}{4a^2}-\frac{c}{a}=\frac{b^2-4ac}{4a^2}$$
$$u=\pm\sqrt{\frac{b^2-4ac}{4a^2}}=\pm\frac{\sqrt{b^2-4ac}}{2a}$$
We now have the solution in terms of a different variable, so let's substitute back:
$$x=u+v=v+u=\frac{-b}{2a}\pm\frac{\sqrt{b^2-4ac}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Similar technique is used on the cubic formula to eliminate the squared term.
A: Youre not asking "what if $3uv + b=0$" and "what if $3uv+b\ne 0$". Youre not understanding what is going on here. 
You are SETTING the equality $3uv+b=0$. You are defining this relationship to be true. Up until this point, $u$ and $v$ were arbitrary variables, independent of one another, and there was an infinite number of $(u,v)$ pairs that satisfied any $x$.  Now they are not arbitrarily defined. Now they are implicitly defined in terms of one another.  One variable is arbitrary and the other is computable from the former.  We went from two arbitrary variables to one arbitrary variable and an equation that relates the two appropriately.  We now only have one pair $(u,v)$ for any given $x$.
The writer of the blog used the phrase "if" when talking about the equality. But they are mistaken... misleading... or they are themselves misled.  Dont worry about the conditional because that is an over complication that is completely unnecessary.
