The plastic constant is solution of :
$$x^3=x+1\tag{1}$$
Considered over the reals.
Why iterating would give other solutions, since using : $x=x^3-1$ is a reversible function and so doesn't add extraneous solutions when substituted in $(1)$?
Then $$(x^3-1)^3=x^3$$ is a degree 9 polynomial.
Could it be that several triples $\{a,b,c\}$ are factorizing the first polynomial, $x^3=x+1$ ?
No. The nine roots of the equation $$ \big(x^3-1\big)^3=x^3 $$ are the three roots $\ r_1, r_2, r_3\ $ of the equation $\ x^3=x+1\ $ (only one of which is real) multiplied by the cube roots of unity, $\ 1, \frac{-1+i\sqrt{3}}{2},\frac{-1-i\sqrt{3}}{2},\ $. Of those nine roots, $\ r_1,r_2,r_3\ $ are the only ones that satisfy the original equation, $\ x^3=x+1\ $.