The standard equations for Du Val singularities are:
$$ A_{n}: x^2+y^2+z^{n+1} =0 \ (n \geq 1) $$ and $$ D_{n}: x^2+y^2z+z^{n-1}=0 \ (n \geq 4) $$
So I wonder how to construct the isomorphism between $A_3$ and $D_3$ ?, Here $D_3 $ is $x^2+y^2z+z^2=0$.
Also, $D_4$ is always written as $x^2+y^3+z^3=0$, why it is isomorphic to $x^2+y^2z+z^3=0$?
For $D_3\cong A_3$, if $x^2+y^2z+z^2=0$, just take $a=x, \ b=z+\frac{1}{2}y^2, \ c=\frac{i+1}{2}y\ $ ,$\ $ you can see $a^2+b^2+c^4=0$.
This idea is from $(y^2+z)^2 = y^4+2y^2z+z^2$
For two forms of $D_4$, if $x^2+y^2z+z^3=0$, take $a=x,\ b=2^{-1/3}z+3^{-1/2}\cdot 2^{-1/3}y,\ c=2^{-1/3}z-3^{-1/2}\cdot 2^{-1/3}y$ , $\ $ you can see that $a^2+b^3+c^3=0$.
And this idea is from the equation $(x+y)^3+(x-y)^3=2x^3+6xy^2$