Derivation of Frey equation from FLT I understand, on  a layman's level, Fray's motivation to write an elliptic equation corresponding to an assumed solution to FLT. My question is, how technically is Frey's equation derived? 
$1.$ FLT : Contradiction. There exist integers $A,B,C,n>2$ such that $A^n + B^n = C^n$
$2.$  Frey: $y^2 = x(x-A^n)(x+B^n)$
$3.$  Elliptic curve $y^2= x^3 + ax + b$  
$4.$  The curve is not modular. => not elliptic => FLT  proved. 
I need is the algebra to go from 1. to 3. above.
Assume Numbers A,B,C such that A^n + B^n = C^n contradicting FLT.
$y^2 = x^3 + ax + b$ an elliptic curve in Weirstrass form.
I am guessing A and B go unchanged into x(x-A^n)(x+B^n) 
Is that correct ?
Expanding 2. I get a term in x^2 , there is none in 3. This is my problem.
What does a $2$ dimensional graph of the Frey curve look like ? 
 A: The algebra to go from 1) to 3) is given in Cremona Chapter III
For clarity here, I will go from 3) to 1) for the general case first.
Basically, $y^2=x^3-27c_4x-54c_6$ following Cremona's change of variables.
$c_4=b_2^2-24b_4$
$c_6=b_2^3+36b_2b_4-216b_6$
$b_2=a_1^2+4a_2$
$b_4=2a_4+a_1a_3$
$b_6=a_3^2+4a_6$
for the Weierstrass equation:
$y^2+a_1xy+a_3y=x^3+a_2x^2+a_4x+a_6$
Next, we write the Frey equation:
$y^2=x(x-a^n)(x+b^n)$ which when expanded is
$y^2=x^3-a^px^2+b^px_2-a^pb^px$
Now you can easily see that for the Frey curve:
$a_1=0$
$a_3=0$
$a_2=b^p-a^p$
$a_4=-a^pb^p$
$a_6=0$
Plug these values in to get $c_4$ and $c_6$ for this Frey curve:
$c_4=16(a^{2p}+b^pa^p+b^{2p})$
$c_6=-32(a^p-b^p)(2a^{2p}-13b^pa^p+2b^{2p})$
Plug these into Cremona's equation and we get the rather unwieldy:
$y^2=x^3-432(a^{2p}+b^pa^p+b^{2p})x+1728(a^p-b^p)(2a^{2p}-13b^pa^p+2b^{2p})$
which is the original Frey curve in 1) in the form you asked for in 3).
Of course, you don't need the curve in this form to prove FLT. Ribet's level lowering of the conductor of the curve shows that the conductor is $2$ and there is no modular elliptic curve of level $2$. $11$ is the first level for a modular elliptic curve. The key is not the Weierstrass manipulation you were seeking.  The key is that the discriminant of the original equation in 1) has an exponent due to FLT's signature $(p,p,p)$.
