Sums, Products, and Powers!

Using the identity $x^{3} + y^{3} = (x + y)(x^{2} - xy + y^{2})$ with $x = a^{2}$ and $y = b^{2}$ we find that

$a^{6} + b^{6} = (a^{2} + b^{2})(a^{4} - a^{2}b^{2} + b^{4}) = (a^{2} + b^{2})((a^{2} + b^{2})^{2} - 3(ab)^{2}).$

Now using the given information we have that $a^{2} + b^{2} = (a + b)^{2} - 2ab = 6^{2} - 2*4 = 28$ , and so

$\dfrac{a^{6} + b^{6}}{16} = \dfrac{(a^{2} + b^{2})((a^{2} + b^{2})^{2} - 3(ab)^{2})}{16} = \dfrac{28*(28^{2} - 3*4^{2})}{16} = \boxed{1288}.$