I was going over twists over elliptic curves. I came over this question. Which refers to this lecture notes, on page $43$ in the second proof, the author mentioned
For simplicity we are going to assume $char\ K \ne 2, 3$ and, in particular, that $E$ and $E′$ can both be put in short Weierstrass form. So suppose we have:
$E_1 : y^2 = x^3 + Ax + B$, and
$E_2 : y^2 = x^3 + A_1x + B_1$.
Then if $j(E_1)\ =\ j(E_2)$, we have $\frac{A^3} {4A^2 + 27B^2} = \frac{A_1^3} {4A_1^3 + 27B_1^2 .}$ Clearing denominators and using $char\ K \ne 3$ we get
$A^3B_1^2= A_1^3B^2$
I did not understand the last step, using cross-multiplication, we get $A^3(4A_1^2+27B_1^2)=A_1^3(4A^2+27B^2)$ from this how can we arrive at $A^3B_1^2= A_1^3B^2$
Also when $j(E_1)=1728, \implies A\ne 0,b=0 $, then do we have to take $u=(\frac{A}{A_1})^{1/4}$ Isn't all the curves with $B=0$ isomorphic? Thank you for your help.