Algebra conundrum

We have that $a^{2} = 4\sqrt{3} + 7.$ To find $a,$ we start by forming the equation

$\sqrt{4\sqrt{3} + 7} = m + n\sqrt{3},$

where $m,n$ are integers. Squaring both sides yields

$4\sqrt{3} + 7 = m^{2} + 3n^{2} + 2mn\sqrt{3}.$

Equating respective rational and irrational components yields that

$mn = 2, m^{2} + 3n^{2} = 7 \Longrightarrow m^{2} + \dfrac{12}{m^{2}} - 7 = 0$

$\Longrightarrow m^{4} - 7m^{2} + 12 = 0 \Longrightarrow (m^{2} - 3)(m^{2} - 4) = 0.$

Now since we require that $m,n$ be integers and $a \gt 0,$ we see that $m = 2, n = 1.$ (We could have done this by simple observation, but it's informative to outline the general process.)

Thus $a = 2 + \sqrt{3}$ and $a^{2} = 7 + 4\sqrt{3}$

$\Longrightarrow a^{3} = (2 + \sqrt{3})(7 + 4\sqrt{3}) = 26 + 15\sqrt{3}.$

Now since $(26 + 15\sqrt{3})(26 - 15\sqrt{3}) = 1$ we have that

$\dfrac{a^{6} - 1}{a^{3}} = \dfrac{(26 + 15\sqrt{3})^{2} - 1}{26 + 15\sqrt{3}} \times \dfrac{26 - 15\sqrt{3}}{26 - 15\sqrt{3}} =$

$((26 + 15\sqrt{3})^{2} - 1)(26 - 15\sqrt{3}) = (26 + 15\sqrt{3}) - (26 - 15\sqrt{3}) = \boxed{30\sqrt{3}}.$