2. Zadatak

Kvadriranjem jednakosti $a-b=3$ dobivamo

$$a^2-2ab+b^2=9,$$

pa je $a^2+b^2=9+2ab=11$. Zato vrijedi

$$\frac1{a^3}-\frac1{b^3} =\frac{b^3-a^3}{(ab)^3} =\frac{(b-a)(b^2+ab+a^2)}{(ab)^3} =-3(11+1)=-36.$$

Nadalje,

$$\frac1{a^6}+\frac1{b^6} =\left(\frac1{a^3}-\frac1{b^3}\right)^2+\frac2{(ab)^3} =(-36)^2+2=1298.$$