= Solution
A <rational map> $V\dashrightarrow\mathbb P^m$ is a morphism on a nonempty open subset of $V$, with two descriptions identified if they agree on a dense open subset. It can be represented there by rational homogeneous coordinate functions not all zero. A point is regular for the map if it extends as a <morphism> on an open neighbourhood of that point.
For the given <projective Cremona transformation>, the three products vanish simultaneously exactly at the three coordinate vertices. Away from them they define a <regular map>. At $(1:0:0)$ consider $[1:t:st]$ as $t\to0$, with $s\ne0$. Its image tends to $[0:s:1]$, which depends on $s$. No regular extension could have these different limiting values. Algebraically these curves lie in its domain for $t\ne0$ and a regular extension must give the same specialization at their common vertex, so the obstruction also works over any infinite characteristic-zero field. The other vertices follow by cyclic symmetry. On the torus $X_0X_1X_2\ne0$, applying the map twice gives
$$
[X_0^2X_1X_2:X_0X_1^2X_2:X_0X_1X_2^2]=[X_0:X_1:X_2].
$$
It is therefore a <birational map> and is its own rational inverse.
Use coordinates $[A:B:C]$ on the target. Substitution in the quintic gives
$$
F(BC,AC,AB)=A^2B^2C^2\,Q(A,B,C),\qquad Q=AC^3+A^3B+B^3C.
$$
Thus on the torus its image is the plane quartic $Q=0$. This quartic is nonsingular even over the algebraic closure. Its partial derivatives are $C^3+3A^2B$, $A^3+3B^2C$ and $3AC^2+B^3$. If one coordinate vanishes at a common zero of these partials, all vanish. If none vanishes, multiplying the three equations gives $A^3B^3C^3=-27A^3B^3C^3$, impossible in characteristic zero. Hence there is no projective singular point.
A smooth plane curve is geometrically irreducible: if it had two positive-degree components, <Bézout theorem> over the algebraic closure would give an intersection point, and the product equation has zero gradient there; a repeated factor also makes the gradient vanish along its component. Therefore this quartic is irreducible. Its torus part is dense and is identified by the <projective Cremona transformation> with the torus part of $V$. No coordinate line is a component of $V$, since restricting $F$ to each coordinate line leaves a nonzero monomial. Consequently every component meets the torus, and its torus part is already irreducible. \b[$V$ is irreducible and birational to the nonsingular quartic $AC^3+A^3B+B^3C=0$.]
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