elliptic curve ${X^3+Y^3=AZ^3}$ consider the elliptic curve $X^3+Y^3=AZ^3$ and $A$ in $K*$ with $O=(1,-1,0)$. show the $j$ invariant of this elliptic curves is $0$. (part d of Silverman exercise page 104 Q3.3 ) 
I can compute the $j$ invariant of Weierstrass form of elliptic curves but I don't know how to change this elliptic to Weierstrass form and compute $j$ invariant. I search a lot about curves but unfortunately, I can't find!
 A: The Handbook of Elliptic Curves by Ian Connell is a wonderful reference.
http://webs.ucm.es/BUCM/mat/doc8354.pdf
(...after a quick search.)
In section [1.4 Cubic to Weierstrass: Nagell's algorithm] we have the steps of the algorithm that can be applied for any (non-singular) cubic to put it in Weierstraß form. (Characteristic not two, three.)
Then section 1.4.1 gives an application on Selmer curves:

Proposition 1.4.1: Consider the Selmer curve
$$au^3+bv^3=c\ ,\qquad abc\ne 0\ ,$$
over some field $K$ of characteristic $\ne 2,3$. Then (after permuting variables) assume
$$
\theta=\sqrt[3]{c/b}\in K\ .
$$
Then the given Selmer curve is birationally equivalent to the curve
$$y^2=x^3-432a^2b^2c^2$$
under the mutually inverse transformations
$$
\begin{aligned}
u &=-\frac{6b\theta^2x}{y-36abc}\ ,\\
v &=\theta\frac{y+36abc}{y-36abc}\ ,\\[2mm]
x &=-\frac{12ab\theta^2u}{v-\theta}\ ,\\
y &=36abc\frac{v+\theta}{v-\theta}\ .
\end{aligned} 
$$

In our case, we can for instance cyclically permute the given equation,
rewrite it like
$$
-AZ^3+X^3=-Y^3\ ,
$$
consider $a=-A$, $b=1$, $c=-1$, so that $\theta=\sqrt[3]{c/b}=-1\in\Bbb Q$, then use the transform in the affine space corresponding to $Y=1$:
$$
\begin{aligned}
Z &=\frac{6x}{y-36A}\ ,\\
X &=-\frac{y+36A}{y-36A}\ ,\ .
\end{aligned} 
$$
and introducing in $-AZ^3+X^3=-1$ we get
$$
-A\left(\frac{6x}{y-36A}\right)^3
-\left(\frac{y+36A}{y-36A}\right)^3
=-1\ .
$$
Now multiply with the common denominator, and reshape equivalently as:
$$
\begin{aligned}
-A\cdot 6^3x^3
&
-y^3
-3\cdot 6^2Ay^2
-3\cdot 6^4A^2y
-6^6A^3
\\
=
&
-y^3
+3\cdot 6^2Ay^2
-3\cdot 6^4A^2y
+6^6A^3\ .
\end{aligned}
$$
The cancellations can now be followed with bare eyes.
A: There's already an answer, but here's how I justify the translations:
If we set $X = U-V, Y = U+V$ we can eliminate one of the cubes:
$$
X^3+Y^3 = 2U^3+ 6UV^2 \\
X^3 + Y^3 - AZ^3 = 2U^3+ 6UV^2 - AZ^3 = 0
$$
And dividing by $U^3$ gives
$$
2 + 6\left(\frac{V}{U}\right)^2 - A\left(\frac{Z}{U}\right)^3 = 0
$$
Setting $\frac{Z}{U} = \frac{n}{6A}, \frac{V}{U} = \frac{m}{6^2A}$ and multiplying across by $6^3A^2$ we get:
$$
432A^2 + m^2 = n^3
$$
which is Weierstrass normal form. I know this isn't the "by-the-book" method, but it's linear rational substitutions only, so you should be able to work it backwards to find the transformation from $(n,m)$ to $(X,Y,Z)$.
