Inversion in a plane; prove $P\left(\Delta A'B'C'\right)=\left[\frac{R^2}{(|OA||OB||OC|)^{\frac{1}{3}}}\right]^4P\left(\Delta ABC\right)$ 
Let $\mathcal I_O^R: M\setminus \{O\}\to M\setminus\{O\}$ be an inversion in a plane $M$ and $\mathcal I_O^R(A)=A',\mathcal I_O^R(B)=B',\mathcal I_O^R(C)=C'$. Prove that:
  $$P\left(\Delta A'B'C'\right)=\left[\frac{R^2}{(|OA||OB||OC|)^{\frac{1}{3}}}\right]^4P\left(\Delta ABC\right)$$
  Where $P(\ldots)$ denotes the area of a triangle.


My attempt:
I wanted to write a complete and concise proof so I decided to return to the beginning and collect the required theory in one place.
Description of construction:
Let $k(O,R)$ be the inversion circle and let $T\in M\setminus\{O\}$ be a point inside the circle, $d(O, T)\lt R$.
Then draw lines $OT$ and $p\perp OT$ s.t. $T\equiv p\cap OT$. Let $D\in k$ be one intersection point of the circle $k$ and line $p$. Then draw a tangent line $t$ of $k$ passing through $D$. The intersection point of $p$ and $OT$ is $T'$.

As seen in the picture, $$\Delta OTD\sim\Delta ODT'\implies \frac{|OT|}{R}=\frac{R}{|OT|}\implies |OT||OT'|=R^2$$ 
If $T$ was outside the circle $k$, we would only follow the procedure backwards which shows $I_O^R$ is involutory.

Now let $I_O^R(A)=A',I_O^R(B)=B'$.
$\implies|OA||OA'|=|OB||OB'|\implies\frac{|OA|}{|OB|}=\frac{|OA'|}{|OB'|}$
Also: $\measuredangle BOA=\measuredangle B'OA'$
$\implies\Delta ABO\sim\Delta A'B'O\implies \frac{|A'B'|}{AB|}=\frac{|OA'|}{|OB|}=\frac{R^2}{|OA||OB|}=\alpha$
$$\implies P(\Delta A'B'O)=\alpha^2 P(\Delta ABO)=\left(\alpha^3\right)^{\frac{2}{3}}P(\Delta ABO)$$


My question:
Is there any elegant method I could use to get this expression or do I have to plug everything into Heron's formula:
$$\left(\frac{R^2}{|OA||OB|}\cdot\frac{R^2}{|OB||OC|}\cdot\frac{R^2}{|OA||OC|}\right)^{\frac{2}{3}}\;?$$

When I was drawing my picture, I first wanted one vertex of $\Delta ABC$ to lie inside the circle and the other two to lie outside in the plane, but that didn't look like conformal mapping, so I drew these triangles instead:

Thank you in advance!
 A: It is false. If $A,B,C$ are distinct points on a line not passing through $O$, inversion at $O$ brings them to distinct non-collinear points, contradicting the claim.
A: The given formula is false for more than one reason, there is already an accepted answer giving a quick reason, but i started an answer trying to give a correct formula, similar to the posted one, draw some pictures, maybe the following is useful.

For reasons that will be transparent from the following computations, i will change some the notations, still using standard notations though.
For a triangle $\Delta ABC$ with sides $a,b,c$, circumcenter $O$, circumcircle radius $R=R_{\Delta ABC}$, and area $S=[ABC]$ we have the known formula
$$
abc=4RS\ ,
$$
which is usable in an avatar of a formula relating the areas from the OP.
I will denote by $\Omega$ the center of the inversion $\mathcal I$, and the power of this inversion will be $k^2$. For a point $X$ we will denote the result of applying $\mathcal I$ on a point $X$, also by $X'$, instead of the more elaborate $\mathcal I(X)$.
So the points $A,B,C$ are mapped to points $A',B',C'$, and it is useful to denote by $O_1$ (which is not $O'$) the circumcenter of the triangle $\Delta A'B'C'$, and by $a',b',c';R'$ the lengths of its sides, and its circum-radius. Then we have the formulas involving the needed areas, 
and for their proportion:
$$
\begin{aligned}[]
4R\;[ABC] &= abc\ ,\\
4R'\;[A'B'C'] &= a'b'c'\ ,\\
\frac{[A'B'C']}{[ABC]}
&=\frac{a'b'c'}{abc}\cdot\frac R{R'}\ .
\end{aligned}
$$
Let us draw a picture.

In case $\Omega$ is "outside" the two circles, we can use tangents from $\Omega $ to them. But in general, we better use the centers line, as above.
Consider the points of intersection $T_1,T_2,T_2',T_1'$ of the two circumcircles $(O)=(ABC)$ and $(O_1)=(A'B'C')$ with the line through $\Omega,O,O_1$. There is an invariance relation:
wiki link : cross-ration invariance for inversion
which shows for the distances $AB$, $BC$, $CA$ ,$2R=T_1T_2$ what they become after inversion. We will use a similar argument, but this is the structural essence.
The triangles $\Delta \Omega AB$ and $\Delta \Omega B'A'$ are similar (with this order of their vertices), and the similarity amounts to the equality of proportions:
$$
\begin{aligned}
\frac{c'}c
&=\frac{A'B'}{BA}
=\frac{\Omega A'}{\Omega B}
=\frac{\Omega B'}{\Omega A}
=
\left(
\frac{\Omega A'}{\Omega B}
\cdot
\frac{\Omega B'}{\Omega A}
\right)^{1/2}\ .
\\[3mm]
&\qquad\text{ Taking the product $(a'/a)(b'/b)(c'/c)$...}
\\[3mm]
\frac{a'b'c'}{abc}
&=
\left(
\frac{\Omega A'}{\Omega A}
\cdot
\frac{\Omega B'}{\Omega B}
\right)^{1/2}
\left(
\frac{\Omega B'}{\Omega B}
\cdot
\frac{\Omega C'}{\Omega C}
\right)^{1/2}
\left(
\frac{\Omega C'}{\Omega C}
\cdot
\frac{\Omega A'}{\Omega A}
\right)^{1/2}
\\
&=
\frac{\Omega A'}{\Omega A}
\cdot
\frac{\Omega B'}{\Omega B}
\cdot
\frac{\Omega C'}{\Omega C}
\\
&=
\frac{k^2}{\Omega A^2}
\cdot
\frac{k^2}{\Omega B^2}
\cdot
\frac{k^2}{\Omega C^2}\ .
\\[3mm]
&\qquad\text{ The proportion of $R,R'$ is:}\\
&\qquad\text{ (same argument, in limit, degenerated case)}
\\
\frac{R}{R'}
&=
\frac{2R}{2R'}
=
\frac{T_1T_2}{T_1'T_2'}
\\
&=
\left(
\frac{\Omega T_1}{\Omega T_1'}
\cdot
\frac{\Omega T_2}{\Omega T_2'}
\right)^{1/2}
=
\left(
\frac{\Omega T_1^2}{\Omega T_1\cdot\Omega T_1'}
\cdot
\frac{\Omega T_2^2}{\Omega T_2\cdot\Omega T_2'}
\right)^{1/2}
\\
&=
\frac{\Omega T_1\cdot\Omega T_2}{k^2}\ .
\\[3mm]
&\qquad\text{ All together:}
\\[3mm]
\frac{[A'B'C']}{[ABC]}
&=\frac{a'b'c'}{abc}\cdot\frac R{R'}
\\
&=
\frac{k^2}{\Omega A^2}
\cdot
\frac{k^2}{\Omega B^2}
\cdot
\frac{k^2}{\Omega C^2}
\cdot 
\frac{\Omega T_1\cdot\Omega T_2}{k^2}
\\
&=
\frac{k^2}{\Omega A^2}
\cdot
\frac{k^2}{\Omega B^2}
\cdot
\frac{k^2}{\Omega C^2}
\cdot 
\frac{\operatorname{Power}(\Omega,(O))}{k^2}
\ .
\end{aligned}
$$
So it is not enough to know only the distances from $\Omega$ to the vertices $A,B,C$, and the inversion factor, we still need to know something about the "position" of $\Omega$ w.r.t. the circle $(O)=(ABC)$, e.g. its power.
