That's hard...maybe

$x^3-3x^2-3x-1=0\\ x^3=3x^2+3x+1\\ 2x^3=x^3+3x^2+3x+1\\ 2x^3=(x+1)^3\\ \sqrt[3]{2}x=x+1\\ (\sqrt[3]{2}-1)x=1\\ x=\frac{1}{\sqrt[3]{2}-1} \quad \Rightarrow \quad x=\frac{1}{\sqrt[3]{2}-1}\cdot\frac{\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1}}{\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1}}=\sqrt[3]{4}+\sqrt[3]{2}+\sqrt[3]{1}\\ a+b+c=7$

Let's use the method of Cardano:

Make the substitution $x=y+1$ in order to delete the quadratic coefficient:

$(y+1)^3-3(y+1)^2-3(y+1)-1=0 \Longrightarrow y^3-6y-6=0$

Now, compare the coefficients of the equation in $y$ with the identity $(a+b)^3-3ab(a+b)-(a^3+b^3)=0$ . An equation system will be formed:

$y=a+b$

$3ab=6$

$a^3+b^3=6$

Simplify the second equation and cube both sides: $(ab)^3=8$

With $(ab)^3=8$ and $a^3+b^3=6$ we are motivated to find an equation with roots $a^3$ and $b^3$ because of Vieta's formulas. Let's make a equation in $z$ : $(z-a^3)(z-b^3)=0 \Longrightarrow z^2-(a^3+b^3)z+(ab)^3=0 \Longrightarrow z^2-6z+8=0$ That is factored eeasily as $(z-2)(z-4)=0$ . Hence, $a^3=2$ and $b^3=4$ .

So, $a=\sqrt[3]{2}$ and $b=\sqrt[3]{4}$ taking the real values.

Finally, $y=\sqrt[3]{2}+\sqrt[3]{4}$ and $x=\sqrt[3]{2}+\sqrt[3]{4}+1$ .

$x=\sqrt[3]{2}+\sqrt[3]{4}+\sqrt[3]{1}$

The final answer is $2+4+1=\boxed{7}$ .

a, b, c ∈ Z