Can you find the sum of the cubes of the roots?

By Vieta's formula, we have in this equation:

$a+b=17$

$ab=19$

Now, we have to find $a^3+b^3$ , which equals to $(a+b)^3-3ab(a+b)$ . Then:

$a^3+b^3=(17)^3-3(19)(17)$

$a^3+b^3=4913-969$

$a^3+b^3=\boxed{3944}$

a3+b3 = (a+b)[(a=b)2-3ab] =17[289-57]=3944