answer:Let's break down the problem into two parts and provide a detailed solution for each. # Part (a): Prove that ( k = a^3 + b^3 + c^3 ) is not a prime number. 1. **Given**: ( a + b + c = 2pq(p^{30} + q^{30}) ) where ( p > q ) are positive integers. 2. **Observation**: Since ( a, b, c ) are natural numbers, ( a + b + c ) is a natural number. 3. **Sum of Cubes**: We know that ( a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca) ). 4. **Given Condition**: ( a + b + c = 2pq(p^{30} + q^{30}) ). 5. **Evenness**: Since ( a + b + c ) is even, ( k = a^3 + b^3 + c^3 ) must also be even. 6. **Prime Check**: The smallest even prime number is 2. For ( k ) to be 2, we would need ( a^3 + b^3 + c^3 = 2 ). However, this is impossible because ( a, b, c ) are natural numbers and the smallest sum of cubes of natural numbers is ( 1^3 + 1^3 + 1^3 = 3 ). 7. **Conclusion**: Since ( k ) is even and greater than 2, ( k ) cannot be a prime number. # Part (b): Prove that if ( a cdot b cdot c ) is maximum, then ( 1984 ) divides ( k ). 1. **Given**: ( a + b + c = 2pq(p^{30} + q^{30}) ). 2. **Divisibility by 3**: Notice that ( a + b + c ) is divisible by 3 because if ( 3 nmid p, q ), then ( 3 mid p^2 - q^2 mid p^{30} - q^{30} ). Hence, ( a + b + c ) is divisible by 3. 3. **AM-GM Inequality**: To maximize ( a cdot b cdot c ), by the Arithmetic Mean-Geometric Mean Inequality (AM-GM), ( a = b = c ). 4. **Expression for ( a )**: Let ( a = b = c = frac{2pq(p^{30} + q^{30})}{3} ). 5. **Expression for ( k )**: [ k = 3a^3 = 3 left( frac{2pq(p^{30} + q^{30})}{3} right)^3 = frac{8p^3q^3(p^{30} + q^{30})^3}{9} ] 6. **Divisibility by 64**: If at least one of ( p ) or ( q ) is even, ( k ) is visibly divisible by ( 64 ). If both are odd, ( (p^{30} - q^{30})^3 ) has an extra factor of ( 8 ) to complete the factor of ( 64 ). 7. **Divisibility by 31**: If at least one of ( p ) or ( q ) is divisible by 31, so is ( k ). If not, ( (31, pq) = 1 ) and by Fermat's Little Theorem, ( p^{30} equiv q^{30} equiv 1 pmod{31} ), so ( p^{30} - q^{30} equiv 0 pmod{31} ). Thus, ( k ) is divisible by 31. 8. **Conclusion**: Combining these, ( k ) is divisible by ( 64 cdot 31 = 1984 ). The final answer is ( boxed{ k } ) is divisible by ( 1984 ).

answer:(1) The domain of the function f(x) is determined by the condition x^2 - 1 neq 0, which implies that x neq pm 1. Therefore, the domain of f(x) is boxed{{x ,|, x neq pm 1}}. (2) To determine the parity of f(x), we evaluate f(-x) as follows: f(-x) = frac{1}{(-x)^2 - 1} = frac{1}{x^2 - 1} = f(x) Thus, f(x) is an even function since f(-x) = f(x) for all x in the domain of f(x). We can express this as boxed{text{f(x) is even}}. (3) To prove the monotonicity of f(x) on the interval (1, +infty), consider any two points x_1, x_2 in (1, +infty) such that x_1 < x_2. The difference f(x_1) - f(x_2) is given by: f(x_1) - f(x_2) = frac{1}{x_1^2 - 1} - frac{1}{x_2^2 - 1} We can combine the fractions to get a common denominator: = frac{(x_2^2 - 1) - (x_1^2 - 1)}{(x_1^2 - 1)(x_2^2 - 1)} = frac{x_2^2 - x_1^2}{(x_1^2 - 1)(x_2^2 - 1)} = frac{(x_2 + x_1)(x_2 - x_1)}{(x_1^2 - 1)(x_2^2 - 1)} Since x_2 > x_1 > 1, both the numerator (x_2 + x_1)(x_2 - x_1) and the denominator (x_1^2 - 1)(x_2^2 - 1) are positive. Therefore: f(x_1) - f(x_2) = frac{(x_2 + x_1)(x_2 - x_1)}{(x_1^2 - 1)(x_2^2 - 1)} > 0 This inequality implies that f(x_1) > f(x_2), meaning that f(x) is a decreasing function on the interval (1, +infty). Thus, we have boxed{text{f(x) is decreasing on (1, +infty)}}.

answer:To find the greatest possible area of (triangle APC), follow these steps: 1. **Identify the Known Parameters and Calculate Area (triangle ABC)**: Given sides of (triangle ABC) are: [ AB = 10, quad BC = 17, quad CA = 21 ] 2. **Calculate the Semi-Perimeter**: The semi-perimeter (s) is given by: [ s = frac{AB + BC + CA}{2} = frac{10 + 17 + 21}{2} = 24 ] 3. **Apply Heron's Formula to Find the Area of (triangle ABC)**: Heron's formula for the area (K) of a triangle with sides (a, b,) and (c) and semi-perimeter (s) is: [ K = sqrt{s(s - a)(s - b)(s - c)} ] Plugging in the values: [ K = sqrt{24(24 - 10)(24 - 17)(24 - 21)} = sqrt{24 cdot 14 cdot 7 cdot 3} ] Calculating inside the square root: [ 24 cdot 14 = 336,quad 336 cdot 7 = 2352,quad 2352 cdot 3 = 7056 ] So, [ K = sqrt{7056} = 84 ] Thus, the area of (triangle ABC) is: [ [ABC] = 84 ] 4. **Calculate the Altitude from (B) to (CA) (denoted as (h_B))**: The area of (triangle ABC) can also be expressed as: [ [ABC] = frac{1}{2} cdot CA cdot h_B ] Solving for (h_B): [ 84 = frac{1}{2} cdot 21 cdot h_B implies h_B = frac{2 cdot 84}{21} = 8 ] 5. **Position of Point (P)**: Since (P) lies on the circle with diameter (AB), and to maximize the area ([APC]), (P) should be positioned such that it is as far from (AC) as possible. 6. **Express Maximum Distance as Diameter Radius**: Since (P) should be at its maximum distance from (A), it should lie on a point such that the altitude from (P) to (AB) becomes the radius (half of diameter (AB)): [ PM = frac{1}{2} AB = frac{10}{2} = 5 ] 7. **Calculate the Maximum Distance from (P) to Line (AC)**: The maximum distance (PQ) from point (P) to line (AC) is the sum of (PM + MQ), where (MQ) is half the altitude (h_B): [ PQ = PM + MQ = 5 + frac{h_B}{2} = 5 + frac{8}{2} = 5 + 4 = 9 ] 8. **Area Calculation for (triangle APC)**: Thus, the area of triangle (APC), when (PQ) is at its maximum: [ [APC] = frac{1}{2} cdot AC cdot PQ = frac{1}{2} cdot 21 cdot 9 = frac{189}{2} = 94.5 ] # Conclusion: (boxed{94.5})

answer:Let v be the volume of soda in Brand W, and let p be the price of Brand W soda. Therefore, the volume of soda in Brand Z is 1.3v, and the price of Brand Z soda is 0.85p. The unit price of Brand Z soda is calculated by dividing the price by the volume, text{Unit price of Brand Z} = frac{0.85p}{1.3v} = frac{85p}{130v}. The unit price of Brand W soda is simply, text{Unit price of Brand W} = frac{p}{v}. To find the ratio of these unit prices, text{Ratio} = frac{frac{85p}{130v}}{frac{p}{v}} = frac{85p}{130v} times frac{v}{p} = frac{85}{130} = frac{17}{26}. Therefore, the ratio of the unit price of Brand Z soda to the unit price of Brand W soda is boxed{frac{17}{26}}.