Помогите решить домашку. Номер 1.25

Решите задачу:

$a+ \frac{1}{a}=3\\ 1) \ a^{2} + \frac{1}{ a^{2} } \\ (a+ \frac{1}{a})^{2}=3^{2} \\ a^{2}+2+ \frac{1}{a^{2}} =9\\ a^{2}+\frac{1}{a^{2}} =7\\ 2) \ \frac{a^{4}+1}{2a^2} = \frac{a^4}{2a^2}+ \frac{1}{2a^{2}} = \frac{a^{2}}{2}+ \frac{1}{2a^2} \\ a^{2}+\frac{1}{a^{2}} =7 |:2 \\ \frac{a^{2}}{2}+ \frac{1}{2a^{2}} =3,5\\ 3) \ \frac{a^{8}+1}{a^{4}} = \frac{a^{8}}{a^{4}} + \frac{1}{a^{4}} =a^{4}+ \frac{1}{a^{4}} \\ (a^{2}+\frac{1}{a^{2}})^2=7^{2} \\ a^{4}+2+ \frac{1}{a^{4}} =49 \\$
$a^{4}+ \frac{1}{a^{4}} =47 \\ 4) a^{3}+\frac{1}{a^{3}} \\ (a+ \frac{1}{a} )^{3}=3^3\\ a^{3}+ \frac{3a^{2}}{a} + \frac{3a}{a^{2}}+ \frac{1}{a^{3}} }= 27\\ a^{3}+ \frac{1}{a^{3}} }= 27- 3a - \frac{3}{a}\\ (a+ \frac{1}{a} )=3 |*3\\ 3a+ \frac{3}{a} =9\\ 3a=9-\frac{3}{a} \\ a^{3}+ \frac{1}{a^{3}} }= 27- 3a - \frac{3}{a}=27-9+\frac{3}{a} -\frac{3}{a} =18$