answer:Since x - 1 divides both x^2 + ax + b and x^2 + bx + c, substituting x = 1 into both equations must yield zero. Therefore: 1. Substitute x = 1: [ 1 + a + b = 0 ] [ 1 + b + c = 0 ] 2. Simplify these equations: [ a + b = -1 ] [ b + c = -1 ] We know from these that a = c. 3. We also have the LCM of these polynomials: [ x^2 + ax + b = (x - 1)(x + b + 1) ] [ x^2 + bx + c = (x - 1)(x + c + 1) ] Given a = c, both can be written as (x - 1)(x + a + 1). As LCM = (x - 1)^2(x + a + 1), this must be equal to given polynomial: [ (x - 1)^2(x + a + 1) = x^3 - 6x^2 + 11x - 6 ] 4. Solving the given LCM: Expanding the obtained LCM expression and comparing coefficients, we get: [ (x - 1)^2(x + a + 1) = x^3 - 2x^2 + x - x^2 + 2x - 1 + a(x^2 - 2x + 1) ] [ = x^3 - (3 + a)x^2 + (3 + 2a)x + (-1 + a) ] Equating coefficients with x^3 - 6x^2 + 11x - 6: [ -3 - a = -6 ] [ a = 3 ] [ 3 + 2a = 11 ] [ 3 + 6 = 11 ] (consistent) 5. Calculate a + b + c given b = c - 1, and c = a: [ a + b + c = 3 + 2 + 3 = boxed{8} ]

answer:The function is given as f(x)=x^{ frac {1}{3}}+ log _{ frac {1}{3}}x. To find f(27), substitute x with 27 in the function: f(27)=27^{ frac {1}{3}}+ log _{ frac {1}{3}}27 Now, let's calculate each term separately: 1. Calculate 27^{ frac {1}{3}}: The cube root of 27 is 3. So, 27^{ frac {1}{3}} = 3. 2. Calculate log _{ frac {1}{3}}27: Since log_a b = frac{log_c b}{log_c a} for any valid base c, we can rewrite log _{ frac {1}{3}}27 as frac{log 27}{log frac{1}{3}}. Using the change of base formula, we can simplify it as follows: frac{log 27}{log frac{1}{3}} = frac{log 3^3}{log 3^{-1}} = frac{3 log 3}{-1 log 3} = -3 Now, add both terms: f(27) = 3 + (-3) = 0 Therefore, the correct answer is: boxed{D: 0}

answer:To find the sum of the digits of the two-thousand-and-first "remarkable" number, we need to understand what it means for a natural number to be "remarkable." According to the problem, a natural number is called "remarkable" if it is the smallest among natural numbers with the same sum of digits. 1. Let's denote the sum of the digits of our "remarkable" number as ( S ). 2. We need to find the two-thousand-and-first smallest number that is the smallest number with a specific sum of digits. Since each "remarkable" number has a unique sum of digits (no two "remarkable" numbers share the same sum of digits), the sum ( S ) for the 2001st "remarkable" number must be 2001. Conclusion: [ boxed{2001} ]

answer:Let's call the weight of the new person W. When the new person replaces the one weighing 75 kg, the total weight of the 6 persons increases by 4.5 kg each. This means the total weight increase for all 6 persons is 6 * 4.5 kg. Total weight increase = 6 * 4.5 kg Total weight increase = 27 kg This increase in total weight is due to the new person's weight being more than the weight of the person they replaced. So, the weight of the new person (W) is equal to the weight of the person they replaced (75 kg) plus the total weight increase (27 kg). W = 75 kg + 27 kg W = 102 kg Therefore, the weight of the new person might be boxed{102} kg.