Integer part of $\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$ Find the value of the following infinite series:
$$\left\lfloor\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}\right\rfloor$$
Now, my doubt is whether it's​ $2$ or it's​ $3$. I'm not sure if it just converges to $3$ but not actually reaches it or if it completely attains the value of $3$. 
I would to like to see the proof (formal) as well instead of just intuitions.

set 
$$x =\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$$
then
$$x^3 = 24+\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$$
$$x^3 = 24 + x$$
Shooting for the rational root theorem, consider the divisors of $24$. 2 does not satisfy the equation, clearly as $8 \neq 26$. However, $3^3 = 24 + 3$, so
$$3 =\sqrt[3]{24+\sqrt[3]{24+\sqrt[3]{24+\cdots}}}$$
Dividing the cubic by $x-3$ yields $x^2 + 3x + 8$ which has no real roots as $9 - 32 = -23 < 0$, so that's our only possible solution.
