Writing $3.8473221018630726$ in the form $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$. I attempted the following question on Brilliant which has to do with finding roots of a cubic polynomial. I was successful in finding what the only real root is but I am facing a problem rewriting the root in the sought expression.

The equation $x^3-3x^2-3x-1=0$  has exactly one real solution that can be written in the form $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$. What is the value of $a+b+c$?

I've found the value of $x$ to be equal to $3.8473221018630726$ by the method of depressing the cubic. Any hints to proceed are appreciated.
Edit:
Would the fact that $x=\dfrac{2}{\sqrt[3]{4}}+\sqrt[3]{4}+1=\dfrac{2}{\sqrt[3]{2}}+\sqrt[3]{2}+1\approx3.8473221018630726$ help somewhere in determining $a, b$ and $c$?

From the numerical solution (which is pretty "small"), you can try to check a few small values for $a,b,c$... which I did:
$$3.8473221018630726=\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}$$
EDIT: Actually this had to be solved in the following way (for one real solution):
$$x^3 - 3x^2 - 3x - 1 = 0$$ 
$$2x^3 - (x+1)^3 = 0$$
$$2x^3 = (x+1)^3$$
$$\sqrt[3]2x = x+1$$
$$x = \frac1{\sqrt[3]2-1}$$
$$x = \frac1{\sqrt[3]2-1}\frac{\sqrt[3]{2^2}+\sqrt[3]{2}+1}{\sqrt[3]{2^2}+\sqrt[3]{2}+1}=\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]1$$
