For a quadratic equation in the form \(ax^2 + bx + c = 0\), the roots are given by the formula \(x = \frac{(-b ± \sqrt{(b^2 - 4ac)}) }{ (2a)}.\) Given that the coefficients a, b, and c are rational, and one root is the irrational number \(2 + \sqrt{3}\), the other root must be its conjugate, \(2 - \sqrt{3}\).
The sum of the roots is \(\frac{(-b) }{ a} = (2 + \sqrt3) + (2 - \sqrt3) = 4.\)
The product of the roots is \(\frac{c }{ a} = (2 + \sqrt3)(2 - \sqrt3) = 4 - 3 = 1.\)
The quadratic equation is actually in the form \(ax^2 - bx + c = 0.\)
Therefore, \(\frac{b }{ a} = 4\) and \(\frac{c }{ a} = 1\).
It is also given that \(b = c^3.\)
Substituting \(b = c^3\) into \(\frac{b }{ a} = 4\), we get \(\frac{(c^3) }{ a} = 4\), which implies \(c^3 = 4a\).
Combining \(\frac{c }{ a} = 1\) (which means \(c = a\)) with \(c^3 = 4a\), we substitute \(a = c\) to get \(c^3 = 4c\). Rearranging this, we have \(c^3 - 4c = 0\), so \(c(c^2 - 4) = 0\). This yields \(c = 0\), \(c = 2\), or \(c = -2\). Since \(c\) is a coefficient of a quadratic equation with rational roots and the roots are \(2 \pm \sqrt{3}\), \(c\) cannot be 0. Thus, \(c = \pm 2\).
If \(c = \pm 2\), then \(a = c = \pm 2\).
Finally, the absolute value of \(a\) is \(|a| = 2.\)