Помогите, пожалуйста, решить задачу по математике. Буду очень благодарен!

Решите задачу:

$a - \frac{1}{a} = \frac{2}{3};\\\\ a) \ a^2 + \frac{1}{a^2} = (a^2 - 2 + \frac{1}{a^2}) + 2 = (a - \frac{1}{a})^2 + 2 = \frac{4}{9} + 2 = 2\frac{4}{9}\\\\$

$b) \ a^3 - \frac{1}{a^3} = (a - \frac{1}{a})(a^2 + 1 + \frac{1}{a^2}) = \frac{2}{3}(a^2 - 2 + \frac{1}{a^2} + 3) =\\\\= \frac{2}{3}([a - \frac{1}{a}]^2 + 3) = \frac{2}{3}[a - \frac{1}{a}]^2 + 2 =\frac{2}{3}*\frac{4}{9} + 2 = \frac{8}{27} + 2= 2\frac{8}{27}$

$c) \ \frac{a^{12}+1}{a^6} = a^6 + \frac{1}{a^6} = [(a^3)^2 - 2 + (\frac{1}{a^3})^2] + 2 = [a^3 -\frac{1}{a^3}]^2 + 2 =\\\\= [2\frac{8}{27}]^2 + 2 = [2 + \frac{8}{27}]^2 + 2 = 4 + \frac{32}{27} + \frac{64}{729} + 2 = 7 + \frac{5}{27} + \frac{64}{729} = \\\\= 7\frac{5*27 + 64}{729} = 7\frac{199}{729}$