How to solve $x+3+\sqrt[3]{(x+1)(x^2-x+2)+\sqrt[3]{x^3+x+1}}+\sqrt[3]{x^3+x+1}=0$ solve the equalition
$$x+3+\sqrt[3]{(x+1)(x^2-x+2)+\sqrt[3]{x^3+x+1}}+\sqrt[3]{x^3+x+1}=0$$
I have seen some methods, for this problem.
my idea:
let $\sqrt[3]{x^3+x+1}=y,$ then
$\sqrt[3]{y^3+y+1}+y=-x-3$
and following have nice methods? thank you everyone.

Let $y=\sqrt[3]{x^3+x+1}$ and $z=\sqrt[3]{y^3+y+1}$, then we translate the original equation into$$x+y+z+3=0$$ by following the procedure in the original post. We also have, by direct calculation:
$$z^3=y^3+y+1=x^3+x+1+y+1$$ so that $$z^3+z+1=x^3+(x+y+z+3)=x^3$$  So we have the suggestive equations$$y^3=x^3+x+1$$$$z^3=y^3+y+1$$$$x^3=z^3+z+1$$
From which we would immediately guess $x=y=z$ and $=-1$ follows immediately.

Suppose the three values are not equal and wlog that $y\gt x$, with $x$ being the least value, then (using the first of the three equations) $$(y-x)(x^2+xy+y^2)=x+1$$ The left-hand side is the product of two positive factors* so we have  $x\gt -1$ and ($x$ being the least value, and values being unequal) $x+y+z\gt -3$, which is a contradiction.

For * note $$x^2+xy+y^2=(x+\frac y2)^2+\frac {3y^2}4$$
