Let $a,$ $b,$ $c$ be the roots of $3x^3 - 3x^2 + 11x - 8 = 0.$  Find $ab + ac + bc.$
By Vieta's formulas, $ab + ac + bc = \boxed{\frac{11}{3}}.$
\frac{11}{3}
Level 2
Intermediate Algebra
Problem: Let $a,$ $b,$ $c$ be the roots of $3x^3 - 3x^2 + 11x - 8 = 0.$  Find $ab + ac + bc.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To find the value of \( ab + ac + bc \), we can utilize Vieta's formulas, which relate the coefficients of a polynomial to sums and products of its roots. For a cubic polynomial of the form \( ax^3 + bx^2 + cx + d = 0 \), if \( r_1, r_2, r_3 \) are the roots, then:

1. The sum of the roots \( r_1 + r_2 + r_3 = -\frac{b}{a} \).
2. The sum of the products of the roots taken two at a time \( r_1r_2 + r_1r_3 + r_2r_3 = \frac{c}{a} \).
3. The product of the roots \( r_1r_2r_3 = -\frac{d}{a} \).

In our case, the polynomial is \( 3x^3 - 3x^2 + 11x - 8 \). Here, \( a = 3 \), \( b = -3 \), \( c = 11 \), and \( d = -8 \).

Applying Vieta's formulas:

1. The sum of the roots \( a + b + c = -\frac{-3}{3} = 1 \).
2. The sum of the products of the roots taken two at a time \( ab + ac + bc = \frac{11}{3} \).
3. The product of the roots \( abc = -\frac{-8}{3} = \frac{8}{3} \).

Since we are specifically interested in \( ab + ac + bc \), we directly use the second relation from Vieta's formulas. Thus, we conclude that the value of \( ab + ac + bc \) is \( \frac{11}{3} \).<|END_OF_LATENT|> The final answer is $\frac{11}{3}$.