answer：Okay, so I have this problem about traffic lights at an intersection. The red light lasts for 30 seconds, the yellow light for 5 seconds, and the green light for 40 seconds. I need to find the probability that when I arrive at the intersection, I encounter a green light. Hmm, probability... I think this has to do with the duration of each light compared to the total cycle of the traffic light.First, let me try to visualize the traffic light cycle. It goes red, then yellow, then green, and then repeats. So, the total time for one full cycle is the sum of the durations of red, yellow, and green lights. That would be 30 seconds for red, plus 5 seconds for yellow, plus 40 seconds for green. Let me add those up: 30 + 5 is 35, and 35 + 40 is 75. So, the total cycle time is 75 seconds.Now, the probability of encountering a green light when arriving at the intersection should be the proportion of time the light is green compared to the total cycle time. That makes sense because if the green light is on for a longer duration, there's a higher chance of arriving during that time. So, the green light is on for 40 seconds out of the total 75 seconds.To find the probability, I can set up a fraction where the numerator is the duration of the green light, and the denominator is the total cycle time. So, that would be 40 seconds divided by 75 seconds. Let me write that as a fraction: 40/75.Now, I should simplify this fraction to its lowest terms. Both 40 and 75 are divisible by 5. Dividing both numerator and denominator by 5, I get 8/15. So, the simplified fraction is 8/15.Wait, let me double-check my calculations. 30 + 5 + 40 is indeed 75. And 40 divided by 75 simplifies to 8/15. Yeah, that seems right. So, the probability of encountering a green light is 8/15.I think that's it. It's all about the proportion of time each light is on relative to the entire cycle. Since the green light is on for a significant portion of the cycle, the probability is higher than for red or yellow. But in this case, the numbers work out neatly to 8/15.

answer：Okay, so I need to find all polynomials ( P(x) ) that satisfy the equation ( x P(x-1) = (x-2) P(x) ) for all real numbers ( x ). Hmm, let me try to break this down step by step.First, I notice that the equation involves ( P(x) ) and ( P(x-1) ), which suggests that it's a functional equation relating the polynomial at different points. Maybe I can substitute specific values of ( x ) to find some properties of ( P(x) ).Let me try substituting ( x = 0 ). Plugging that into the equation gives:[0 cdot P(-1) = (-2) cdot P(0)]Simplifying, the left side is 0, and the right side is (-2 P(0)). So, ( 0 = -2 P(0) ), which implies that ( P(0) = 0 ). Okay, so ( x = 0 ) is a root of the polynomial ( P(x) ).Next, let me try ( x = 2 ). Substituting ( x = 2 ) into the equation:[2 P(1) = 0 cdot P(2)]The right side is 0, so ( 2 P(1) = 0 ), which means ( P(1) = 0 ). So, ( x = 1 ) is also a root of ( P(x) ).Now, I know that ( P(0) = 0 ) and ( P(1) = 0 ), which means that ( P(x) ) must have factors of ( x ) and ( (x - 1) ). Therefore, I can express ( P(x) ) as:[P(x) = x(x - 1) Q(x)]where ( Q(x) ) is another polynomial. Let me substitute this back into the original equation to see what happens.Substituting ( P(x) = x(x - 1) Q(x) ) into ( x P(x - 1) = (x - 2) P(x) ):[x cdot (x - 1)(x - 2) Q(x - 1) = (x - 2) cdot x(x - 1) Q(x)]Wait, let me make sure I expand that correctly. Let's compute each side step by step.Left side: ( x P(x - 1) )[x cdot (x - 1)(x - 2) Q(x - 1)]Right side: ( (x - 2) P(x) )[(x - 2) cdot x(x - 1) Q(x)]So, both sides have ( x(x - 1)(x - 2) ) as a factor. Let me cancel those out, assuming ( x neq 0, 1, 2 ) to avoid division by zero. Then, we get:[Q(x - 1) = Q(x)]Hmm, this equation suggests that ( Q(x) ) is a polynomial that is invariant under a shift of 1. The only polynomials that satisfy ( Q(x) = Q(x - 1) ) for all ( x ) are constant polynomials. Because if a polynomial is periodic, it must be constant. Otherwise, the degree would have to be zero.So, ( Q(x) ) must be a constant, say ( a ). Therefore, substituting back into ( P(x) ):[P(x) = a x(x - 1)]So, ( P(x) ) is a quadratic polynomial with leading coefficient ( a ).Let me verify this solution by plugging it back into the original equation. Let ( P(x) = a x(x - 1) ).Compute ( x P(x - 1) ):[x cdot a (x - 1)(x - 2)]Which simplifies to:[a x(x - 1)(x - 2)]Compute ( (x - 2) P(x) ):[(x - 2) cdot a x(x - 1)]Which also simplifies to:[a x(x - 1)(x - 2)]So, both sides are equal, which confirms that ( P(x) = a x(x - 1) ) satisfies the equation.Wait, but I should also consider if there are any other possible solutions. For example, could ( P(x) ) be the zero polynomial? If ( P(x) = 0 ), then both sides of the equation become zero, which is valid. So, the zero polynomial is also a solution, which is included in the general solution when ( a = 0 ).Therefore, all solutions are scalar multiples of ( x(x - 1) ), including the zero polynomial.I think that's it. I don't see any other possibilities because the functional equation forces ( P(x) ) to have roots at 0 and 1, and then the remaining factor must be a constant. So, the general solution is ( P(x) = a x(x - 1) ) where ( a ) is a constant.**Final Answer**The polynomials satisfying the equation are all constant multiples of ( x(x - 1) ), so the solution is boxed{P(x) = a(x^2 - x)} where ( a ) is a constant.**Final Answer**The polynomials ( P(x) ) that satisfy the equation are given by boxed{P(x) = a(x^2 - x)} where ( a ) is a constant.

answer：Alright, so I have this problem where I need to find three distinct integers a, b, and c between 0 and 9 such that the three-digit number abc is equal to a! + b! + c!. Hmm, okay, let's break this down step by step.First, I know that n! means the factorial of n, which is the product of all positive integers up to n. For example, 5! is 120 because 1×2×3×4×5 equals 120. Also, by convention, 0! and 1! are both equal to 1. So, that's good to remember.Now, I need to find three distinct digits a, b, and c such that when I put them together as a three-digit number, it equals the sum of their factorials. So, abc = a! + b! + c!. Let's think about what this means.Since abc is a three-digit number, it must be between 100 and 999. That gives me a range to work with. Also, a, b, and c are digits, so they can only be from 0 to 9, and they have to be distinct, meaning no repeats.Let me list out the factorials for digits 0 through 9 to have a clear reference:- 0! = 1- 1! = 1- 2! = 2- 3! = 6- 4! = 24- 5! = 120- 6! = 720- 7! = 5040- 8! = 40320- 9! = 362880Wait a minute, 7!, 8!, and 9! are way too big because they result in four or five-digit numbers. Since our target abc is a three-digit number, including 7, 8, or 9 in a, b, or c would make the sum a! + b! + c! too large. So, I can probably rule out 7, 8, and 9 because their factorials exceed three digits.So, now I'm focusing on digits 0 through 6. Let's list their factorials again for clarity:- 0! = 1- 1! = 1- 2! = 2- 3! = 6- 4! = 24- 5! = 120- 6! = 720Hmm, 6! is 720, which is still a three-digit number. So, 6 might be a candidate because 720 is within our range. But wait, if one of the digits is 6, then the sum a! + b! + c! could potentially be as high as 720 + something, which might push it beyond three digits if the other digits are large. Let's see.But before that, let's think about the maximum possible sum of three factorials. If we take the three largest factorials in our range, which are 6! = 720, 5! = 120, and 4! = 24. Adding them up: 720 + 120 + 24 = 864. That's still a three-digit number. So, actually, 6 is possible because even with 6!, the total sum doesn't exceed three digits.But wait, 6! is 720, which is almost three digits by itself. So, if one of the digits is 6, the other two digits must be small enough so that their factorials don't push the total sum beyond 999. Let's consider that.But before diving into that, maybe it's better to approach this systematically. Let's consider the possible values for a, b, and c, keeping in mind that they must be distinct and between 0 and 6.Since abc is a three-digit number, a cannot be 0. So, a must be between 1 and 6.Let me start by assuming that a is 1. Then, I need to find b and c such that 1bc = 1! + b! + c!.So, 100 + 10b + c = 1 + b! + c!Which simplifies to 10b + c = b! + c! - 99Hmm, that seems a bit complicated. Maybe instead of starting with a = 1, I should look for known numbers that are equal to the sum of their digits' factorials. I remember hearing about something called "factorions," which are numbers that equal the sum of the factorials of their digits. In base 10, there are only a few of these numbers.From what I recall, the factorions in base 10 are 1, 2, 145, and 40585. But 40585 is a five-digit number, so it's outside our range. So, 145 is a three-digit factorion. Let me check that.145: 1! + 4! + 5! = 1 + 24 + 120 = 145. Yes, that works. So, 145 is a factorion. Therefore, a = 1, b = 4, c = 5.Wait, but the problem says to find three distinct integers a, b, and c between 0 and 9. In this case, 1, 4, and 5 are distinct and between 0 and 9, so that should be the solution.But let me make sure there are no other solutions. Let's check if there are any other three-digit factorions.I know 40585 is a five-digit factorion, but are there others in three digits? Let me think.Let me check 135: 1! + 3! + 5! = 1 + 6 + 120 = 127, which is not 135.How about 153: 1! + 5! + 3! = 1 + 120 + 6 = 127, same as above.Wait, 127 is interesting. 1! + 2! + 7! = 1 + 2 + 5040 = 5043, which is way too big. So, 127 doesn't work.What about 222: 2! + 2! + 2! = 2 + 2 + 2 = 6, which is not 222.How about 145: as above, it works.Is there another one? Let's see.Let me try 123: 1! + 2! + 3! = 1 + 2 + 6 = 9, not 123.134: 1! + 3! + 4! = 1 + 6 + 24 = 31, not 134.144: 1! + 4! + 4! = 1 + 24 + 24 = 49, not 144.155: 1! + 5! + 5! = 1 + 120 + 120 = 241, not 155.222: already checked.223: 2! + 2! + 3! = 2 + 2 + 6 = 10, not 223.234: 2! + 3! + 4! = 2 + 6 + 24 = 32, not 234.333: 3! + 3! + 3! = 6 + 6 + 6 = 18, not 333.444: 4! + 4! + 4! = 24 + 24 + 24 = 72, not 444.555: 5! + 5! + 5! = 120 + 120 + 120 = 360, not 555.666: 6! + 6! + 6! = 720 + 720 + 720 = 2160, way too big.Okay, so it seems like 145 is the only three-digit factorion. Therefore, the digits a, b, and c must be 1, 4, and 5.But just to be thorough, let me check if there are any other combinations where a is not 1.Suppose a = 2. Then, the number is 2bc, and we have 200 + 10b + c = 2! + b! + c! = 2 + b! + c!.So, 10b + c = b! + c! - 198.Hmm, that seems tricky because b! + c! needs to be at least 198 + 10b + c. But b and c are digits from 0 to 6, so their factorials are at most 720, but likely much smaller.Wait, let's see. The maximum b! + c! can be is 720 + 120 = 840. So, 840 - 198 = 642. So, 10b + c needs to be 642? But 10b + c is at most 66 (if b=6 and c=6), which is way less than 642. So, this is impossible. Therefore, a cannot be 2.Similarly, if a = 3, then 300 + 10b + c = 6 + b! + c!.So, 10b + c = b! + c! - 294.Again, the right side needs to be at least 294 + 10b + c. But b! + c! is at most 840, so 840 - 294 = 546. So, 10b + c needs to be 546, which is impossible because 10b + c is at most 66. So, a cannot be 3.Similarly, for a = 4:400 + 10b + c = 24 + b! + c!.So, 10b + c = b! + c! - 376.Again, b! + c! needs to be at least 376 + 10b + c. The maximum b! + c! is 840, so 840 - 376 = 464. 10b + c needs to be 464, which is impossible.For a = 5:500 + 10b + c = 120 + b! + c!.So, 10b + c = b! + c! - 380.Again, b! + c! needs to be at least 380 + 10b + c. The maximum is 840, so 840 - 380 = 460. 10b + c needs to be 460, which is impossible.For a = 6:600 + 10b + c = 720 + b! + c!.So, 10b + c = b! + c! - 120.Here, b! + c! needs to be at least 120 + 10b + c. Let's see, the minimum b! + c! is 1 + 1 = 2, but we need it to be at least 120 + 10b + c. Since b and c are digits, 10b + c is at most 66, so 120 + 66 = 186. So, b! + c! needs to be at least 186.Looking at factorials, 5! = 120, 6! = 720. So, if one of b or c is 5, then b! + c! is at least 120 + 1 = 121, which is close. Let's see.If b = 5, then 10b + c = 50 + c. So, 50 + c = 120 + c! - 120 = c!.Wait, that would mean 50 + c = c!.But c is a digit from 0 to 6. Let's check:c = 0: 0! = 1, 50 + 0 = 50 ≠ 1c = 1: 1! = 1, 50 + 1 = 51 ≠ 1c = 2: 2! = 2, 50 + 2 = 52 ≠ 2c = 3: 6, 50 + 3 = 53 ≠ 6c = 4: 24, 50 + 4 = 54 ≠ 24c = 5: 120, 50 + 5 = 55 ≠ 120c = 6: 720, 50 + 6 = 56 ≠ 720So, no solution here.What if b = 6? Then, 10b + c = 60 + c. So, 60 + c = 720 + c! - 120 = 600 + c!.So, 60 + c = 600 + c!.Rearranged: c! = 60 + c - 600 = c - 540.But c! is positive, and c - 540 is negative for c from 0 to 6. So, no solution here either.What if both b and c are 5? Then, b! + c! = 120 + 120 = 240.So, 10b + c = 50 + 5 = 55.But 240 - 120 = 120, which is not equal to 55. So, no.Alternatively, if b = 4 and c = 5:b! + c! = 24 + 120 = 144.So, 10b + c = 40 + 5 = 45.But 144 - 120 = 24, which is not 45.Wait, maybe I'm approaching this wrong. Let's think differently.If a = 6, then 600 + 10b + c = 720 + b! + c!.So, 10b + c = b! + c! + 720 - 600 = b! + c! + 120.But 10b + c is at most 66, while b! + c! + 120 is at least 120 + 1 + 1 = 122. So, 10b + c needs to be at least 122, which is impossible because 10b + c is at most 66. Therefore, no solution when a = 6.So, after checking all possible values for a from 1 to 6, the only solution is when a = 1, b = 4, and c = 5, which gives us the number 145, and indeed 1! + 4! + 5! = 1 + 24 + 120 = 145.Therefore, the three distinct integers are 1, 4, and 5.

answer：Okay, so I've got this problem about Zack and his marbles. Let me try to understand it step by step. The problem says that Zack's number of marbles can be divided equally among three people, leaving Zack with 5. Then, if Zack decides to give his three friends 20 marbles each and keeps five, we need to find out how many marbles he initially had.Alright, first, let's break down the first part: "Zack's number of marbles can be divided equally among three people, leaving Zack with 5." So, if Zack has a certain number of marbles, and when he divides them equally among three people, he's left with 5. That means if he gives the same number of marbles to each of the three people, he still has 5 marbles remaining.Let me think about this. If he divides his marbles equally among three people, that means he's giving away marbles in multiples of 3. So, the total number of marbles he has must be a multiple of 3 plus 5. In other words, if we let M be the total number of marbles Zack has, then M = 3k + 5, where k is the number of marbles each person gets.Now, the second part of the problem says that Zack decides to give his three friends 20 marbles each and keeps five. So, he gives away 20 marbles to each of his three friends and keeps 5 for himself. That means the total number of marbles he gives away is 3 times 20, which is 60 marbles, plus the 5 he keeps, making a total of 65 marbles.Wait a minute, so if M = 65, does that fit with the first part of the problem? Let's check. If M = 65, then according to the first part, M should be equal to 3k + 5. So, 65 = 3k + 5. Solving for k, we subtract 5 from both sides: 60 = 3k. Then, dividing both sides by 3, we get k = 20. That means each person gets 20 marbles, and Zack is left with 5. That matches perfectly with the second part of the problem where he gives each friend 20 marbles and keeps 5.So, it seems like the initial number of marbles Zack had is 65. But let me double-check to make sure I didn't make any mistakes. If Zack has 65 marbles and he divides them equally among three people, each person gets 20 marbles (since 3 times 20 is 60), and Zack is left with 5 marbles. That fits the first condition. Then, when he decides to give his three friends 20 marbles each and keeps five, he gives away 60 marbles and keeps 5, totaling 65 marbles. That matches as well.Hmm, I think I've got it right. Zack initially had 65 marbles.

answer：First, I need to determine how many letters Jack received in the morning. I'll start by defining variables for the number of letters received in the morning (( L_m )) and in the afternoon (( L_a )).From the problem, I know that Jack received 1 more letter in the morning than in the afternoon. This relationship can be expressed as:[L_m = L_a + 1]Additionally, it's given that Jack received 7 letters in the afternoon:[L_a = 7]Now, I can substitute the value of ( L_a ) into the first equation to find ( L_m ):[L_m = 7 + 1 = 8]Therefore, Jack received 8 letters in the morning.

answer：Alright, so I have this problem here where I need to find the maximum value of the expression (frac{2a^3 + 27c - 9ab}{lambda^3}) given some conditions on the roots of a cubic polynomial. Let me try to unpack this step by step.First, the polynomial is (f(x) = x^3 + ax^2 + bx + c), and it has three real roots (x_1, x_2, x_3). The conditions given are:1. (x_2 - x_1 = lambda), where (lambda) is positive.2. (x_3 > frac{1}{2}(x_1 + x_2)).I need to find the maximum value of (frac{2a^3 + 27c - 9ab}{lambda^3}).Okay, so I remember that for a cubic polynomial with roots (x_1, x_2, x_3), Vieta's formulas relate the coefficients (a, b, c) to the roots. Specifically:- (x_1 + x_2 + x_3 = -a)- (x_1x_2 + x_2x_3 + x_3x_1 = b)- (x_1x_2x_3 = -c)So, maybe I can express (a, b, c) in terms of the roots and then substitute them into the expression I need to maximize.Given that (x_2 - x_1 = lambda), it might be helpful to set (x_1 = m - frac{lambda}{2}) and (x_2 = m + frac{lambda}{2}), where (m) is the midpoint between (x_1) and (x_2). That way, the difference between (x_2) and (x_1) is exactly (lambda). Now, the second condition is (x_3 > frac{1}{2}(x_1 + x_2)). Since (x_1 + x_2 = 2m), this simplifies to (x_3 > m). So, let me denote (x_3 = m + t), where (t > 0).So now, my roots are:- (x_1 = m - frac{lambda}{2})- (x_2 = m + frac{lambda}{2})- (x_3 = m + t)Great, now I can express (a, b, c) in terms of (m, lambda, t).Starting with (a):(x_1 + x_2 + x_3 = (m - frac{lambda}{2}) + (m + frac{lambda}{2}) + (m + t) = 3m + t)So, (a = -(3m + t)).Next, (b):(x_1x_2 + x_2x_3 + x_3x_1)Let me compute each term:- (x_1x_2 = (m - frac{lambda}{2})(m + frac{lambda}{2}) = m^2 - frac{lambda^2}{4})- (x_2x_3 = (m + frac{lambda}{2})(m + t) = m^2 + mt + frac{lambda}{2}m + frac{lambda}{2}t)- (x_3x_1 = (m + t)(m - frac{lambda}{2}) = m^2 - frac{lambda}{2}m + mt - frac{lambda}{2}t)Adding them up:(x_1x_2 + x_2x_3 + x_3x_1 = (m^2 - frac{lambda^2}{4}) + (m^2 + mt + frac{lambda}{2}m + frac{lambda}{2}t) + (m^2 - frac{lambda}{2}m + mt - frac{lambda}{2}t))Simplify term by term:- (m^2 + m^2 + m^2 = 3m^2)- (mt + mt = 2mt)- (frac{lambda}{2}m - frac{lambda}{2}m = 0)- (frac{lambda}{2}t - frac{lambda}{2}t = 0)- (-frac{lambda^2}{4})So, altogether:(3m^2 + 2mt - frac{lambda^2}{4})Thus, (b = 3m^2 + 2mt - frac{lambda^2}{4}).Now, (c):(x_1x_2x_3 = (m - frac{lambda}{2})(m + frac{lambda}{2})(m + t))First, compute ((m - frac{lambda}{2})(m + frac{lambda}{2}) = m^2 - frac{lambda^2}{4})Then, multiply by ((m + t)):((m^2 - frac{lambda^2}{4})(m + t) = m^3 + m^2 t - frac{lambda^2}{4}m - frac{lambda^2}{4}t)So, (x_1x_2x_3 = m^3 + m^2 t - frac{lambda^2}{4}m - frac{lambda^2}{4}t)Therefore, (c = -m^3 - m^2 t + frac{lambda^2}{4}m + frac{lambda^2}{4}t)Alright, now I have expressions for (a, b, c) in terms of (m, lambda, t). Let me write them down:- (a = -(3m + t))- (b = 3m^2 + 2mt - frac{lambda^2}{4})- (c = -m^3 - m^2 t + frac{lambda^2}{4}m + frac{lambda^2}{4}t)Now, I need to compute (2a^3 + 27c - 9ab) and then divide by (lambda^3).Let me compute each term step by step.First, compute (2a^3):(a = -(3m + t)), so (a^3 = -(3m + t)^3)Let me expand ((3m + t)^3):((3m + t)^3 = 27m^3 + 27m^2 t + 9m t^2 + t^3)So, (a^3 = -27m^3 - 27m^2 t - 9m t^2 - t^3)Multiply by 2:(2a^3 = -54m^3 - 54m^2 t - 18m t^2 - 2t^3)Next, compute (27c):(c = -m^3 - m^2 t + frac{lambda^2}{4}m + frac{lambda^2}{4}t)Multiply by 27:(27c = -27m^3 - 27m^2 t + frac{27lambda^2}{4}m + frac{27lambda^2}{4}t)Now, compute (9ab):First, (a = -(3m + t)), (b = 3m^2 + 2mt - frac{lambda^2}{4})Multiply them:(ab = -(3m + t)(3m^2 + 2mt - frac{lambda^2}{4}))Let me expand this:Multiply term by term:- (3m * 3m^2 = 9m^3)- (3m * 2mt = 6m^2 t)- (3m * (-frac{lambda^2}{4}) = -frac{3lambda^2}{4}m)- (t * 3m^2 = 3m^2 t)- (t * 2mt = 2m t^2)- (t * (-frac{lambda^2}{4}) = -frac{lambda^2}{4}t)So, adding all together:(9m^3 + 6m^2 t - frac{3lambda^2}{4}m + 3m^2 t + 2m t^2 - frac{lambda^2}{4}t)Combine like terms:- (9m^3)- (6m^2 t + 3m^2 t = 9m^2 t)- (2m t^2)- (-frac{3lambda^2}{4}m - frac{lambda^2}{4}t = -frac{lambda^2}{4}(3m + t))So, (ab = 9m^3 + 9m^2 t + 2m t^2 - frac{lambda^2}{4}(3m + t))But remember, (ab) was multiplied by (-1) because (a = -(3m + t)), so:(ab = -(9m^3 + 9m^2 t + 2m t^2 - frac{lambda^2}{4}(3m + t)))Wait, no, actually, (ab = -(3m + t)(3m^2 + 2mt - frac{lambda^2}{4})), which we expanded to:(ab = -[9m^3 + 9m^2 t + 2m t^2 - frac{lambda^2}{4}(3m + t)])So, (ab = -9m^3 - 9m^2 t - 2m t^2 + frac{lambda^2}{4}(3m + t))Therefore, (9ab = 9*(-9m^3 - 9m^2 t - 2m t^2 + frac{lambda^2}{4}(3m + t)))Compute this:(9ab = -81m^3 - 81m^2 t - 18m t^2 + frac{27lambda^2}{4}m + frac{9lambda^2}{4}t)Wait, hold on, that doesn't seem right. Let me double-check.Wait, no, actually, (ab = -9m^3 - 9m^2 t - 2m t^2 + frac{lambda^2}{4}(3m + t)), so multiplying by 9:(9ab = -81m^3 - 81m^2 t - 18m t^2 + frac{27lambda^2}{4}m + frac{9lambda^2}{4}t)Yes, that's correct.Now, let's put it all together:(2a^3 + 27c - 9ab = (2a^3) + (27c) - (9ab))Substitute the expressions:(2a^3 = -54m^3 - 54m^2 t - 18m t^2 - 2t^3)(27c = -27m^3 - 27m^2 t + frac{27lambda^2}{4}m + frac{27lambda^2}{4}t)(-9ab = 81m^3 + 81m^2 t + 18m t^2 - frac{27lambda^2}{4}m - frac{9lambda^2}{4}t)Now, add them all together:Let me write each term:- From (2a^3): - (-54m^3) - (-54m^2 t) - (-18m t^2) - (-2t^3)- From (27c): - (-27m^3) - (-27m^2 t) - (+frac{27lambda^2}{4}m) - (+frac{27lambda^2}{4}t)- From (-9ab): - (+81m^3) - (+81m^2 t) - (+18m t^2) - (-frac{27lambda^2}{4}m) - (-frac{9lambda^2}{4}t)Now, combine like terms:For (m^3):- (-54m^3 -27m^3 +81m^3 = (-54 -27 +81)m^3 = 0m^3)For (m^2 t):- (-54m^2 t -27m^2 t +81m^2 t = (-54 -27 +81)m^2 t = 0m^2 t)For (m t^2):- (-18m t^2 +18m t^2 = 0m t^2)For (t^3):- (-2t^3)For (m):- (+frac{27lambda^2}{4}m - frac{27lambda^2}{4}m = 0m)For (t):- (+frac{27lambda^2}{4}t - frac{9lambda^2}{4}t = frac{18lambda^2}{4}t = frac{9lambda^2}{2}t)So, putting it all together:(2a^3 + 27c - 9ab = -2t^3 + frac{9lambda^2}{2}t)So, the expression simplifies to:(frac{2a^3 + 27c - 9ab}{lambda^3} = frac{-2t^3 + frac{9lambda^2}{2}t}{lambda^3})Let me factor out (frac{t}{lambda^3}):(frac{t}{lambda^3}(-2t^2 + frac{9lambda^2}{2}))Alternatively, factor out (frac{t}{lambda^3}):(frac{t}{lambda^3}(-2t^2 + frac{9}{2}lambda^2))But perhaps it's better to write it as:(frac{9lambda^2 t - 4t^3}{2lambda^3})Yes, because:(-2t^3 + frac{9}{2}lambda^2 t = frac{9}{2}lambda^2 t - 2t^3)So, dividing by (lambda^3):(frac{frac{9}{2}lambda^2 t - 2t^3}{lambda^3} = frac{9}{2}frac{t}{lambda} - 2left(frac{t}{lambda}right)^3)Let me set (k = frac{t}{lambda}), which is a positive real number since both (t) and (lambda) are positive.So, the expression becomes:(frac{9}{2}k - 2k^3)So, now, I need to maximize the function (f(k) = frac{9}{2}k - 2k^3) with respect to (k > 0).To find the maximum, take the derivative and set it to zero.(f'(k) = frac{9}{2} - 6k^2)Set (f'(k) = 0):(frac{9}{2} - 6k^2 = 0)Solving for (k^2):(6k^2 = frac{9}{2})(k^2 = frac{9}{12} = frac{3}{4})So, (k = sqrt{frac{3}{4}} = frac{sqrt{3}}{2})Since (k > 0), we take the positive root.Now, check if this is a maximum by using the second derivative:(f''(k) = -12k)At (k = frac{sqrt{3}}{2}), (f''(k) = -12 * frac{sqrt{3}}{2} = -6sqrt{3} < 0), which means it's a local maximum.So, the maximum value of (f(k)) is:(fleft(frac{sqrt{3}}{2}right) = frac{9}{2} * frac{sqrt{3}}{2} - 2 * left(frac{sqrt{3}}{2}right)^3)Compute each term:First term: (frac{9}{2} * frac{sqrt{3}}{2} = frac{9sqrt{3}}{4})Second term: (2 * left(frac{sqrt{3}}{2}right)^3 = 2 * frac{3sqrt{3}}{8} = frac{6sqrt{3}}{8} = frac{3sqrt{3}}{4})So, subtracting:(frac{9sqrt{3}}{4} - frac{3sqrt{3}}{4} = frac{6sqrt{3}}{4} = frac{3sqrt{3}}{2})Therefore, the maximum value of (frac{2a^3 + 27c - 9ab}{lambda^3}) is (frac{3sqrt{3}}{2}).I should verify if this maximum is achievable under the given conditions. Since (k = frac{sqrt{3}}{2}), which implies (t = frac{sqrt{3}}{2} lambda). Given that (t > 0), this is valid. So, the maximum is indeed achievable.**Final Answer**The maximum value is (boxed{dfrac{3sqrt{3}}{2}}).