Polynomial coefficients based off of its roots

When we look at $x=\sqrt[3]{2+\sqrt{3}}+\sqrt[3]{2-\sqrt{3}}$ , we don't like the cube roots sitting there. That gives us the idea to cube both sides.

Cubing, we get:

$x^3=2+\sqrt{3}+2-\sqrt{3}+3(\sqrt[3]{(2+\sqrt{3})^2(2-\sqrt{3})}+\sqrt[3]{(2+\sqrt{3})(2-\sqrt{3})^2})$

When we look at those ugly radicals, we see difference of squares: $(2+\sqrt{3})(2-\sqrt{3})=1$ which means those ugly radicals become:

$3(\sqrt[3]{2+\sqrt{3}}+\sqrt[3]{2-\sqrt{3}})$

But we already know $x=\sqrt[3]{2+\sqrt{3}}+\sqrt[3]{2-\sqrt{3}}$ so the equation becomes: $x^3=4+3x$

Putting into standard form and finding the absolute value of the sum of the coefficients, we get $\boxed{7}$