Just substitution right?

One way of solve it is:

$x^2(x-18)+101x-132$

$x^2(6+\sqrt{7}-18)+101x-132$

$x^2(\sqrt{7}-12)+101x-132$

$x((\sqrt{7}-12))(6+\sqrt{7})+101)-132$

$(6+\sqrt{7})(36-6\sqrt{7})-132$

$\boxed{216+36\sqrt{6}-36\sqrt{6}-42-132=42}$

First let's find the minimal polynomial of $x$ :

$x-6=\sqrt{7} \\ x^2-12x+36=7 \\ x^2-12x+29=0$

Now use polynomial long division to obtain:

$x^3-18x^2+101x-132=(x-6)(x^2-12x+29)+42 \\ x^3-18x^2+101x-132=(x-6)(0)+42 \\ x^3-18x^2+101x-132=\boxed{42}$