Key Concept (Vieta's Formulas): For $x^3 + bx^2 + cx + d = 0$ with roots $r_1, r_2, r_3$:
$$r_1+r_2+r_3 = -b, \quad r_1r_2+r_1r_3+r_2r_3 = c, \quad r_1r_2r_3 = -d$$
Also useful: $r_1^2+r_2^2+r_3^2 = (r_1+r_2+r_3)^2 - 2(r_1r_2+r_1r_3+r_2r_3)$.
If $x^3 - 6x^2 + 11x - 6 = 0$ has roots $r_1, r_2, r_3$, find $r_1^2 + r_2^2 + r_3^2$.
Solution: By Vieta's: $\sum r_i = 6$, $\sum r_ir_j = 11$.
Then $\sum r_i^2 = 6^2 - 2(11) = 36 - 22 = 14$.
Answer: 14
The polynomial $x^3 - 9x^2 + 26x - 24 = 0$ has three positive integer roots. What is the sum of the squares of the roots?
29|49|53|77|81
By Vieta's formulas: $r_1+r_2+r_3 = 9$, $\;r_1r_2+r_1r_3+r_2r_3 = 26$, $\;r_1r_2r_3 = 24$.
Test integers whose product is 24 and sum is 9: try $2, 3, 4$.
Check: $2+3+4 = 9$ ✓ and $2{\cdot}3 + 2{\cdot}4 + 3{\cdot}4 = 6+8+12 = 26$ ✓
Sum of squares: $2^2 + 3^2 + 4^2 = 4 + 9 + 36 = \boxed{49}$.