Problem 2 - Entrance Test

From Vieta's formulas for the polynomial P(x) = x^3 - 3x^2 + 2x - 1, we have: 1. Sum of the roots: a + b + c = -(-3)/1 = 3 2. Sum of the products of the roots taken two at a time: ab + bc + ca = 2/1 = 2 3. Product of the roots: abc = -(-1)/1 = 1 We want to find the value of (a+b)(b+c)(c+a). From (1), we can express each term in the product: a + b = 3 - c b + c = 3 - a c + a = 3 - b Substitute these into the expression: (a+b)(b+c)(c+a) = (3-c)(3-a)(3-b). This expression is related to the polynomial P(x). Recall that if a, b, c are the roots of P(x), then P(x) can be written as (x-a)(x-b)(x-c) (assuming the leading coefficient is 1, which it is in this case). So, P(x) = (x-a)(x-b)(x-c). The expression (3-c)(3-a)(3-b) is exactly P(3). Let's calculate P(3): P(3) = 3^3 - 3(3^2) + 2(3) - 1 P(3) = 27 - 3(9) + 6 - 1 P(3) = 27 - 27 + 6 - 1 P(3) = 5. Thus, the value of (a+b)(b+c)(c+a) is 5.