Question 548600
By Vieta's formulas, a+b+c = 2. Hence we are evaluating (2-c)(2-b)(2-a). We could expand this entire expression (since we are guaranteed to obtain useful cyclic sums:


*[tex \LARGE (2-c)(2-b)(2-a) = 8 - 4a - 4b - 4c + 2ab + 2ac + 2bc - abc]


*[tex \LARGE = 8 - 4(a+b+c) + 2(ab + ac + bc) - abc]


However, we have a+b+c = 2, ab+ac+bc = 3, abc = 4 (you can verify these by expanding (x-a)(x-b)(x-c) and equating like terms). Hence, the expression we want is equal to


*[tex \LARGE 8 - 4(2) + 2(3) - 4 = 2]