$a-b=3$
$↔ a=b+3$ $(1)$
$a^2+b^2=8$ $(2)$
Thay $(1)$ vào $(2)$, ta có:
$(b+3)^2+b^2=8$
$↔ 2b^2+6b+1=0$
$↔ \left[ \begin{array}{l}x=\dfrac{-3+\sqrt[]{7}}{2}\\x=\dfrac{-3-\sqrt[]{7}}{2}\end{array} \right.$
$ab=\dfrac{-3+\sqrt[]{7}}{2}.\dfrac{-3-\sqrt[]{7}}{2}=\dfrac{1}{2}$
$a^3+b^3=\Bigg(\dfrac{-3+\sqrt[]{7}}{2}\Bigg)^3+\Bigg(\dfrac{-3-\sqrt[]{7}}{2}\Bigg)^3=-\dfrac{45}{2}$
