answer：First, I need to determine how many tables each person served individually and then sum those amounts to find the total number of tables served by all three.Emily worked the full 9 hours, serving 5 tables each hour. So, Emily served 9 multiplied by 5, which equals 45 tables.Olivia had to attend a meeting and only worked for 7 hours. At 5 tables per hour, Olivia served 7 multiplied by 5, totaling 35 tables.Ava took a longer break and worked for 8 hours. Therefore, Ava served 8 multiplied by 5, which is 40 tables.Adding up the tables served by each person: 45 (Emily) + 35 (Olivia) + 40 (Ava) equals 120 tables in total.

answer：First, I need to determine the total number of cupcakes Carl needs to sell to reach his goal, which is 96 cupcakes over 2 days.Next, I should account for the cupcakes he needs to give to Bonnie. Carl needs to give Bonnie 24 cupcakes as payment for using her storefront.To find out how many cupcakes Carl must sell each day, I'll subtract the cupcakes given to Bonnie from the total goal. This means subtracting 24 from 96, which equals 72 cupcakes.Finally, I'll divide the remaining cupcakes by the number of days, which is 2. Dividing 72 by 2 gives me 36 cupcakes per day.Therefore, Carl must sell 36 cupcakes each day to reach his goal.

answer：First, I need to determine the dimensions of the rectangle. The total length of the wire is 28 cm, which means the perimeter of the rectangle is 28 cm.Let’s denote the width of the rectangle as ( w ) cm. According to the problem, the length is 4 cm more than the width, so the length can be expressed as ( w + 4 ) cm.The formula for the perimeter of a rectangle is ( 2 times (text{length} + text{width}) ). Plugging in the expressions for length and width, the equation becomes:[2 times (w + 4 + w) = 28]Simplifying inside the parentheses:[2 times (2w + 4) = 28]Multiplying:[4w + 8 = 28]Subtracting 8 from both sides:[4w = 20]Dividing by 4:[w = 5]So, the width is 5 cm. The length, being 4 cm more than the width, is:[5 + 4 = 9 text{ cm}]

answer：Okay, so I have this problem about an aquatic product dealer who buys a certain species of freshwater fish at 30 yuan per kilogram. The dealer sells these fish, and the daily sales volume, y, is linearly related to the selling price, x. The selling price ranges from 30 to 60 yuan per kilogram. They gave me a table with two data points: when x is 50 yuan, y is 100 kilograms, and when x is 40 yuan, y is 200 kilograms.Part (1) asks me to find the expression of y in terms of x. Since it's a linear relationship, I know that y can be expressed as y = mx + b, where m is the slope and b is the y-intercept. So, I need to find m and b using the given data points.Let me write down the two equations based on the given points. When x = 50, y = 100:100 = m*50 + b.And when x = 40, y = 200:200 = m*40 + b.Now, I have a system of two equations:1) 50m + b = 1002) 40m + b = 200I can solve this system to find m and b. Let me subtract equation 2 from equation 1 to eliminate b:(50m + b) - (40m + b) = 100 - 200Simplify:10m = -100So, m = -10.Now that I have m, I can plug it back into one of the equations to find b. Let's use equation 1:50*(-10) + b = 100-500 + b = 100Add 500 to both sides:b = 600.So, the equation is y = -10x + 600.Let me double-check with the second point to make sure. If x = 40:y = -10*40 + 600 = -400 + 600 = 200. That's correct. And for x = 50:y = -10*50 + 600 = -500 + 600 = 100. Perfect, that matches the given data.So, part (1) is done. The expression is y = -10x + 600.Moving on to part (2). The dealer's daily sales profit, W, is to be maximized. I need to express W in terms of x and then find the value of x that maximizes it.First, let's recall that profit is calculated as (selling price - cost price) multiplied by the number of units sold. Here, the cost price is 30 yuan per kilogram, and the selling price is x yuan per kilogram. The number of kilograms sold is y, which we already have as a function of x.So, profit W can be expressed as:W = (x - 30) * y.But since y is a function of x, we can substitute y = -10x + 600 into this equation:W = (x - 30)*(-10x + 600).Let me expand this expression:First, multiply x by each term in the second parenthesis:x*(-10x) + x*600 = -10x² + 600x.Then, multiply -30 by each term in the second parenthesis:-30*(-10x) + (-30)*600 = 300x - 18000.Now, combine these two results:-10x² + 600x + 300x - 18000.Combine like terms:-10x² + (600x + 300x) - 18000 = -10x² + 900x - 18000.So, the profit function is W = -10x² + 900x - 18000.Now, this is a quadratic function in terms of x, and since the coefficient of x² is negative (-10), the parabola opens downward, meaning the vertex is the maximum point.To find the value of x that maximizes W, I can use the vertex formula for a parabola. The x-coordinate of the vertex is given by x = -b/(2a), where a and b are coefficients from the quadratic equation ax² + bx + c.In this case, a = -10 and b = 900.So, x = -900/(2*(-10)) = -900/(-20) = 45.So, the selling price x that maximizes the profit is 45 yuan per kilogram.Now, let's find the maximum profit W by plugging x = 45 back into the profit equation:W = -10*(45)² + 900*(45) - 18000.First, compute (45)²: 45*45 = 2025.So, -10*2025 = -20250.Then, 900*45: Let's compute 900*40 = 36,000 and 900*5 = 4,500. So, 36,000 + 4,500 = 40,500.Now, add these together with the last term:-20250 + 40,500 - 18,000.Let me compute step by step:First, -20250 + 40,500 = 20,250.Then, 20,250 - 18,000 = 2,250.So, the maximum profit is 2,250 yuan.Wait, let me verify that calculation again because 2,250 seems a bit low. Let me compute W at x = 45 again.Alternatively, since W = (x - 30)*y, and y = -10x + 600, when x = 45:y = -10*45 + 600 = -450 + 600 = 150 kilograms.So, profit W = (45 - 30)*150 = 15*150 = 2,250 yuan.Yes, that's correct. So, the maximum profit is indeed 2,250 yuan when the selling price is 45 yuan per kilogram.Just to make sure, let me check another point near 45 to see if the profit is indeed lower. Let's try x = 44:y = -10*44 + 600 = -440 + 600 = 160.Profit W = (44 - 30)*160 = 14*160 = 2,240 yuan, which is less than 2,250.Similarly, x = 46:y = -10*46 + 600 = -460 + 600 = 140.Profit W = (46 - 30)*140 = 16*140 = 2,240 yuan, also less than 2,250.Hence, 45 yuan per kilogram gives the maximum profit of 2,250 yuan.So, summarizing:1) The expression for y in terms of x is y = -10x + 600.2) The selling price x that maximizes the daily sales profit is 45 yuan per kilogram, and the maximum profit is 2,250 yuan.**Final Answer**(1) The expression of ( y ) in terms of ( x ) is (boxed{y = -10x + 600}).(2) The selling price ( x ) that maximizes the daily sales profit is (boxed{45}) yuan per kilogram, and the maximum daily sales profit is (boxed{2250}) yuan.

answer：Alright, so I have this problem where I need to find all pairs of positive integers (a, b) such that both a³ + 6ab + 1 and b³ + 6ab + 1 are perfect cubes. Hmm, okay. Let me try to unpack this step by step.First, let me write down what I know. I need both expressions, a³ + 6ab + 1 and b³ + 6ab + 1, to be perfect cubes. Let me denote them as:C₁ = a³ + 6ab + 1C₂ = b³ + 6ab + 1So, both C₁ and C₂ must be perfect cubes. Let me think about what that means. If C₁ is a perfect cube, then there exists some integer x such that C₁ = x³. Similarly, C₂ = y³ for some integer y.So, we have:a³ + 6ab + 1 = x³b³ + 6ab + 1 = y³Now, since a and b are positive integers, x and y must also be positive integers greater than a and b respectively, because a³ + 6ab + 1 is definitely larger than a³, and similarly for b³ + 6ab + 1.Wait, is that necessarily true? Let me check. Suppose a is 1, then 1 + 6b + 1 = 6b + 2. If b is 1, that's 8, which is 2³. So, x could be 2 in that case, which is larger than a=1. Similarly, for b=1, 1 + 6a + 1 = 6a + 2. If a=1, that's 8 again, so y=2. So, in that case, x and y are larger than a and b. But what if a or b is larger? Let me see.Suppose a=2, then C₁ = 8 + 12b + 1 = 12b + 9. If b=1, that's 21, which is not a cube. If b=2, that's 33, not a cube. b=3, 45, not a cube. b=4, 57, not a cube. b=5, 69, not a cube. Hmm, maybe a=2 doesn't work. Let me try a=3.C₁ = 27 + 18b + 1 = 18b + 28. Let's see for b=1: 46, not a cube. b=2: 64, which is 4³. So, x=4. So, in this case, x=4 is larger than a=3. Similarly, C₂ for b=2 would be 8 + 36 + 1 = 45, which is not a cube. So, that doesn't work. Hmm.Wait, so in the case where a=3 and b=2, C₁ is a cube, but C₂ isn't. So, that pair doesn't satisfy both conditions. So, maybe I need to find a and b such that both expressions are cubes.Let me think about symmetry here. The problem is symmetric in a and b, so if (a, b) is a solution, then (b, a) is also a solution. That might be helpful.So, maybe I can assume without loss of generality that a ≤ b, find all such pairs, and then include their symmetric counterparts.Let me try small values of a and b to see if I can find any solutions.Starting with a=1:C₁ = 1 + 6b + 1 = 6b + 2C₂ = b³ + 6b + 1We need both 6b + 2 and b³ + 6b + 1 to be cubes.Let me see for b=1:C₁=8, which is 2³, and C₂=8, which is 2³. So, (1,1) is a solution.b=2:C₁=14, not a cube.b=3:C₁=20, not a cube.b=4:C₁=26, not a cube.b=5:C₁=32, not a cube.b=6:C₁=38, not a cube.b=7:C₁=44, not a cube.b=8:C₁=50, not a cube.b=9:C₁=56, not a cube.b=10:C₁=62, not a cube.Hmm, seems like for a=1, only b=1 gives a cube for C₁. Let me check C₂ for b=1: 1 + 6 + 1=8, which is 2³. So, yes, (1,1) works.Now, let's try a=2:C₁=8 + 12b + 1=12b + 9C₂=b³ + 12b + 1We need both 12b + 9 and b³ + 12b + 1 to be cubes.Let's try b=1:C₁=21, not a cube.b=2:C₁=33, not a cube.b=3:C₁=45, not a cube.b=4:C₁=57, not a cube.b=5:C₁=69, not a cube.b=6:C₁=81, which is 3⁴, not a cube.Wait, 81 is 3⁴, which is not a cube. So, no.b=7:C₁=93, not a cube.b=8:C₁=105, not a cube.b=9:C₁=117, not a cube.b=10:C₁=129, not a cube.Hmm, not seeing any cubes here. Maybe a=2 doesn't work.a=3:C₁=27 + 18b +1=18b +28C₂=b³ + 18b +1Looking for both to be cubes.b=1:C₁=46, not a cube.b=2:C₁=64, which is 4³. So, x=4.C₂=8 + 36 +1=45, not a cube.So, no.b=3:C₁=18*3 +28=54 +28=82, not a cube.b=4:C₁=72 +28=100, not a cube.b=5:C₁=90 +28=118, not a cube.b=6:C₁=108 +28=136, not a cube.b=7:C₁=126 +28=154, not a cube.b=8:C₁=144 +28=172, not a cube.b=9:C₁=162 +28=190, not a cube.b=10:C₁=180 +28=208, not a cube.Hmm, no luck with a=3.a=4:C₁=64 + 24b +1=24b +65C₂=b³ +24b +1Looking for both to be cubes.b=1:C₁=89, not a cube.b=2:C₁=24*2 +65=48 +65=113, not a cube.b=3:C₁=72 +65=137, not a cube.b=4:C₁=96 +65=161, not a cube.b=5:C₁=120 +65=185, not a cube.b=6:C₁=144 +65=209, not a cube.b=7:C₁=168 +65=233, not a cube.b=8:C₁=192 +65=257, not a cube.b=9:C₁=216 +65=281, not a cube.b=10:C₁=240 +65=305, not a cube.Nope, nothing here.a=5:C₁=125 + 30b +1=30b +126C₂=b³ +30b +1Looking for cubes.b=1:C₁=156, not a cube.b=2:C₁=60 +126=186, not a cube.b=3:C₁=90 +126=216, which is 6³. So, x=6.C₂=27 +90 +1=118, not a cube.So, no.b=4:C₁=120 +126=246, not a cube.b=5:C₁=150 +126=276, not a cube.b=6:C₁=180 +126=306, not a cube.b=7:C₁=210 +126=336, not a cube.b=8:C₁=240 +126=366, not a cube.b=9:C₁=270 +126=396, not a cube.b=10:C₁=300 +126=426, not a cube.Nope.a=6:C₁=216 + 36b +1=36b +217C₂=b³ +36b +1Looking for cubes.b=1:C₁=36 +217=253, not a cube.b=2:C₁=72 +217=289, which is 17², not a cube.b=3:C₁=108 +217=325, not a cube.b=4:C₁=144 +217=361, which is 19², not a cube.b=5:C₁=180 +217=397, not a cube.b=6:C₁=216 +217=433, not a cube.b=7:C₁=252 +217=469, not a cube.b=8:C₁=288 +217=505, not a cube.b=9:C₁=324 +217=541, not a cube.b=10:C₁=360 +217=577, not a cube.Hmm, nothing here either.a=7:C₁=343 +42b +1=42b +344C₂=b³ +42b +1Looking for cubes.b=1:C₁=42 +344=386, not a cube.b=2:C₁=84 +344=428, not a cube.b=3:C₁=126 +344=470, not a cube.b=4:C₁=168 +344=512, which is 8³. So, x=8.C₂=64 + 168 +1=233, not a cube.So, no.b=5:C₁=210 +344=554, not a cube.b=6:C₁=252 +344=596, not a cube.b=7:C₁=294 +344=638, not a cube.b=8:C₁=336 +344=680, not a cube.b=9:C₁=378 +344=722, not a cube.b=10:C₁=420 +344=764, not a cube.No luck.a=8:C₁=512 +48b +1=48b +513C₂=b³ +48b +1Looking for cubes.b=1:C₁=48 +513=561, not a cube.b=2:C₁=96 +513=609, not a cube.b=3:C₁=144 +513=657, not a cube.b=4:C₁=192 +513=705, not a cube.b=5:C₁=240 +513=753, not a cube.b=6:C₁=288 +513=801, not a cube.b=7:C₁=336 +513=849, not a cube.b=8:C₁=384 +513=897, not a cube.b=9:C₁=432 +513=945, not a cube.b=10:C₁=480 +513=993, not a cube.Hmm, nothing here either.a=9:C₁=729 +54b +1=54b +730C₂=b³ +54b +1Looking for cubes.b=1:C₁=54 +730=784, which is 28², not a cube.b=2:C₁=108 +730=838, not a cube.b=3:C₁=162 +730=892, not a cube.b=4:C₁=216 +730=946, not a cube.b=5:C₁=270 +730=1000, which is 10³. So, x=10.C₂=125 +270 +1=396, not a cube.So, no.b=6:C₁=324 +730=1054, not a cube.b=7:C₁=378 +730=1108, not a cube.b=8:C₁=432 +730=1162, not a cube.b=9:C₁=486 +730=1216, not a cube.b=10:C₁=540 +730=1270, not a cube.Nope.a=10:C₁=1000 +60b +1=60b +1001C₂=b³ +60b +1Looking for cubes.b=1:C₁=60 +1001=1061, not a cube.b=2:C₁=120 +1001=1121, not a cube.b=3:C₁=180 +1001=1181, not a cube.b=4:C₁=240 +1001=1241, not a cube.b=5:C₁=300 +1001=1301, not a cube.b=6:C₁=360 +1001=1361, not a cube.b=7:C₁=420 +1001=1421, not a cube.b=8:C₁=480 +1001=1481, not a cube.b=9:C₁=540 +1001=1541, not a cube.b=10:C₁=600 +1001=1601, not a cube.Hmm, nothing here either.So, up to a=10, the only solution I found is (1,1). Maybe I should check higher a and b, but this might take a while. Alternatively, perhaps there's a smarter way to approach this problem rather than brute-forcing.Let me think about the equations again:a³ + 6ab + 1 = x³b³ + 6ab + 1 = y³Let me subtract the two equations:(a³ + 6ab + 1) - (b³ + 6ab + 1) = x³ - y³Simplify:a³ - b³ = x³ - y³Factor both sides:(a - b)(a² + ab + b²) = (x - y)(x² + xy + y²)Hmm, interesting. So, the product of (a - b) and (a² + ab + b²) equals the product of (x - y) and (x² + xy + y²).Since a and b are positive integers, a - b could be positive, negative, or zero. But since we're dealing with positive integers, if a > b, then a - b is positive, and if a < b, it's negative. However, since we can assume without loss of generality that a ≤ b, let's suppose a ≤ b, so a - b is non-positive. But since we have a product equal to another product, maybe we can find some relationship between the factors.Alternatively, perhaps we can consider that both x and y are close to a and b respectively. Let me think about how much 6ab +1 can affect the cube.Let me consider that a³ + 6ab +1 is close to a³, so x is probably a + k for some small k. Similarly, y is probably b + m for some small m.Let me try to approximate x and y.Suppose x = a + k, then x³ = (a + k)³ = a³ + 3a²k + 3ak² + k³So, a³ + 6ab +1 = a³ + 3a²k + 3ak² + k³Subtract a³:6ab +1 = 3a²k + 3ak² + k³Similarly, for y = b + m:b³ + 6ab +1 = (b + m)³ = b³ + 3b²m + 3bm² + m³Subtract b³:6ab +1 = 3b²m + 3bm² + m³So, now we have two equations:6ab +1 = 3a²k + 3ak² + k³ ...(1)6ab +1 = 3b²m + 3bm² + m³ ...(2)Since both equal 6ab +1, we can set them equal to each other:3a²k + 3ak² + k³ = 3b²m + 3bm² + m³Hmm, this seems complicated, but maybe we can find some relationship between k and m.Alternatively, perhaps k and m are small integers, like 1 or 2, so let's try k=1.If k=1:Equation (1): 6ab +1 = 3a² + 3a +1Simplify:6ab = 3a² + 3aDivide both sides by 3a (since a is positive integer, a ≠ 0):2b = a +1So, b = (a +1)/2Since b must be an integer, a must be odd. Let me denote a = 2n +1, so b = (2n +1 +1)/2 = n +1.So, a = 2n +1, b = n +1.Now, let's plug this into equation (2):6ab +1 = 3b²m + 3bm² + m³First, compute 6ab:6ab = 6*(2n +1)*(n +1) = 6*(2n² +3n +1) = 12n² +18n +6So, 6ab +1 = 12n² +18n +7Now, equation (2):12n² +18n +7 = 3b²m + 3bm² + m³But b = n +1, so:12n² +18n +7 = 3(n +1)²m + 3(n +1)m² + m³Let me expand the right-hand side:3(n² + 2n +1)m + 3(n +1)m² + m³= 3n²m +6nm +3m +3nm² +3m² +m³So, the equation becomes:12n² +18n +7 = 3n²m +6nm +3m +3nm² +3m² +m³Hmm, this seems a bit messy, but maybe we can try small values of m.Let me try m=1:RHS = 3n²*1 +6n*1 +3*1 +3n*1² +3*1² +1³= 3n² +6n +3 +3n +3 +1= 3n² +9n +7Compare to LHS: 12n² +18n +7So, 3n² +9n +7 = 12n² +18n +7Subtract LHS:0 = 9n² +9nWhich implies 9n(n +1)=0Since n is a non-negative integer (as a=2n+1 is positive), n=0.So, n=0, which gives a=1, b=1.So, that's the solution we found earlier.Now, let's try m=2:RHS = 3n²*2 +6n*2 +3*2 +3n*2² +3*2² +2³=6n² +12n +6 +12n +12 +8=6n² +24n +26Set equal to LHS:12n² +18n +7So:6n² +24n +26 =12n² +18n +7Bring everything to one side:-6n² +6n +19=0Multiply by -1:6n² -6n -19=0Discriminant: 36 + 456=492Square root of 492 is about 22.18, not integer. So, no integer solution.m=3:RHS=3n²*3 +6n*3 +3*3 +3n*3² +3*3² +3³=9n² +18n +9 +27n +27 +27=9n² +45n +63Set equal to LHS:12n² +18n +7So:9n² +45n +63 =12n² +18n +7Bring everything to one side:-3n² +27n +56=0Multiply by -1:3n² -27n -56=0Discriminant: 729 + 672=1401Square root of 1401 is about 37.44, not integer. No solution.m=4:RHS=3n²*4 +6n*4 +3*4 +3n*4² +3*4² +4³=12n² +24n +12 +48n +48 +64=12n² +72n +124Set equal to LHS:12n² +18n +7So:12n² +72n +124 =12n² +18n +7Subtract LHS:0 +54n +117=054n +117=0n= -117/54= -2.166..., not positive integer. So, no solution.m=5:RHS=3n²*5 +6n*5 +3*5 +3n*5² +3*5² +5³=15n² +30n +15 +75n +75 +125=15n² +105n +215Set equal to LHS:12n² +18n +7So:15n² +105n +215 =12n² +18n +7Bring everything to one side:3n² +87n +208=0Discriminant: 7569 - 2496=5073Square root of 5073 is about 71.23, not integer.So, no solution.Hmm, seems like m=1 is the only possible, giving n=0, which is a=1, b=1.So, maybe that's the only solution when k=1.Let me try k=2.So, k=2:Equation (1):6ab +1 =3a²*2 +3a*2² +2³=6a² +12a +8So:6ab +1 =6a² +12a +8Bring all terms to one side:6ab -6a² -12a -7=0Divide both sides by 1 (can't simplify much):6ab =6a² +12a +7Divide both sides by a (a≠0):6b =6a +12 +7/aSince b must be integer, 7/a must be integer, so a divides 7.Possible a:1,7.a=1:6b=6*1 +12 +7/1=6 +12 +7=25So, 6b=25 => b=25/6, not integer.a=7:6b=6*7 +12 +7/7=42 +12 +1=55So, 6b=55 => b=55/6, not integer.So, no solution for k=2.k=3:Equation (1):6ab +1=3a²*3 +3a*3² +3³=9a² +27a +27So:6ab +1=9a² +27a +27Bring all terms to one side:6ab -9a² -27a -26=0Divide by 1:6ab=9a² +27a +26Divide both sides by a:6b=9a +27 +26/aSo, 26/a must be integer, so a divides 26.Possible a:1,2,13,26.a=1:6b=9 +27 +26=62 => b=62/6=31/3, not integer.a=2:6b=18 +27 +13=58 => b=58/6=29/3, not integer.a=13:6b=117 +27 +2=146 => b=146/6=73/3, not integer.a=26:6b=234 +27 +1=262 => b=262/6=131/3, not integer.No solution.k=4:Equation (1):6ab +1=3a²*4 +3a*4² +4³=12a² +48a +64So:6ab +1=12a² +48a +64Bring all terms to one side:6ab -12a² -48a -63=0Divide by 3:2ab -4a² -16a -21=0Rearrange:2ab=4a² +16a +21Divide both sides by a:2b=4a +16 +21/aSo, 21/a must be integer, so a divides 21.Possible a:1,3,7,21.a=1:2b=4 +16 +21=41 => b=41/2, not integer.a=3:2b=12 +16 +7=35 => b=35/2, not integer.a=7:2b=28 +16 +3=47 => b=47/2, not integer.a=21:2b=84 +16 +1=101 => b=101/2, not integer.No solution.k=5:Equation (1):6ab +1=3a²*5 +3a*5² +5³=15a² +75a +125So:6ab +1=15a² +75a +125Bring all terms to one side:6ab -15a² -75a -124=0Divide by 1:6ab=15a² +75a +124Divide both sides by a:6b=15a +75 +124/aSo, 124/a must be integer, so a divides 124.Possible a:1,2,4,31,62,124.a=1:6b=15 +75 +124=214 => b=214/6=107/3, not integer.a=2:6b=30 +75 +62=167 => b=167/6, not integer.a=4:6b=60 +75 +31=166 => b=166/6=83/3, not integer.a=31:6b=465 +75 +4=544 => b=544/6=272/3, not integer.a=62:6b=930 +75 +2=1007 => b=1007/6, not integer.a=124:6b=1860 +75 +1=1936 => b=1936/6=968/3, not integer.No solution.Hmm, this is getting tedious, but it seems like k=1 is the only possible case that gives a solution, which is a=1, b=1.Alternatively, maybe I should consider that x and y are not necessarily a +k and b +m, but perhaps some other relation.Wait, another approach: Let me consider that a³ + 6ab +1 is a cube. Let me denote this as x³, so:x³ = a³ + 6ab +1Similarly, y³ = b³ + 6ab +1Let me subtract the two equations:x³ - y³ = a³ - b³Which factors as:(x - y)(x² + xy + y²) = (a - b)(a² + ab + b²)Now, since x³ > a³ (since 6ab +1 >0), x > a. Similarly, y > b.But since x³ - y³ = a³ - b³, if a > b, then x³ > y³, so x > y. Similarly, if a < b, then x < y.But since we assumed a ≤ b, then x ≤ y.Wait, but x > a and y > b, so if a ≤ b, then x could be less than or greater than y depending on the relation between a and b.Wait, actually, if a ≤ b, then a³ ≤ b³, so x³ - y³ = a³ - b³ ≤0, so x³ ≤ y³, so x ≤ y.So, x ≤ y.But x > a and y > b, and a ≤ b, so x could be greater than a but less than or equal to y, which is greater than b.Hmm, not sure if that helps.Alternatively, perhaps we can consider that x³ - a³ = 6ab +1 and y³ - b³ =6ab +1.So, x³ - a³ = y³ - b³.Which implies x³ - y³ = a³ - b³.Which is the same as before.Alternatively, perhaps we can write x³ - a³ = y³ - b³.Which is (x - a)(x² + xa + a²) = (y - b)(y² + yb + b²)Hmm, maybe we can set (x - a) = (y - b) and (x² + xa + a²) = (y² + yb + b²). But that might not necessarily hold, but let's suppose that.So, suppose:x - a = y - b = k (some positive integer)andx² + xa + a² = y² + yb + b²So, x = a + k, y = b + kThen, substitute into the second equation:(a + k)² + (a + k)a + a² = (b + k)² + (b + k)b + b²Expand:(a² + 2ak + k²) + (a² + ak) + a² = (b² + 2bk + k²) + (b² + bk) + b²Simplify:3a² + 3ak + k² = 3b² + 3bk + k²Cancel k²:3a² + 3ak = 3b² + 3bkDivide both sides by 3:a² + ak = b² + bkRearrange:a² - b² + ak - bk =0Factor:(a - b)(a + b) + k(a - b)=0Factor out (a - b):(a - b)(a + b + k)=0So, either a - b=0 or a + b + k=0.But a and b are positive integers, so a + b + k cannot be zero. Therefore, a - b=0, so a = b.So, if a = b, then x = a + k, y = a + k, so x = y.But then, from the original equations:x³ = a³ +6a² +1Similarly, y³ = a³ +6a² +1, so same as x³.So, x = y.But then, we have x³ = a³ +6a² +1.So, let me write that:x³ = a³ +6a² +1Let me see if this can be a cube.Let me try small a:a=1:x³=1 +6 +1=8=2³, so x=2.So, a=1, b=1 is a solution.a=2:x³=8 +24 +1=33, not a cube.a=3:x³=27 +54 +1=82, not a cube.a=4:x³=64 +96 +1=161, not a cube.a=5:x³=125 +150 +1=276, not a cube.a=6:x³=216 +216 +1=433, not a cube.a=7:x³=343 +294 +1=638, not a cube.a=8:x³=512 +384 +1=897, not a cube.a=9:x³=729 +486 +1=1216, not a cube.a=10:x³=1000 +600 +1=1601, not a cube.So, only a=1 gives a cube here.Therefore, the only solution when a=b is (1,1).But earlier, when we considered k=1, we also got a=1, b=1 as the only solution.So, perhaps (1,1) is the only solution where a=b.But what about when a ≠ b? Earlier, when a=3, b=2, C₁ was a cube but C₂ wasn't. Similarly, when a=2, b=3, C₂ was a cube but C₁ wasn't.Wait, let me check a=2, b=3:C₁=8 + 36 +1=45, not a cube.C₂=27 +36 +1=64=4³. So, only C₂ is a cube.Similarly, a=3, b=2:C₁=27 +36 +1=64=4³C₂=8 +36 +1=45, not a cube.So, in these cases, only one of them is a cube.But maybe there are other pairs where both are cubes.Wait, let me think about the case when a=0, but a must be positive integer, so a=0 is invalid.Alternatively, maybe a=1, b=2:C₁=1 +12 +1=14, not a cube.C₂=8 +12 +1=21, not a cube.a=1, b=3:C₁=1 +18 +1=20, not a cube.C₂=27 +18 +1=46, not a cube.a=1, b=4:C₁=1 +24 +1=26, not a cube.C₂=64 +24 +1=89, not a cube.a=1, b=5:C₁=1 +30 +1=32, not a cube.C₂=125 +30 +1=156, not a cube.a=1, b=6:C₁=1 +36 +1=38, not a cube.C₂=216 +36 +1=253, not a cube.a=1, b=7:C₁=1 +42 +1=44, not a cube.C₂=343 +42 +1=386, not a cube.a=1, b=8:C₁=1 +48 +1=50, not a cube.C₂=512 +48 +1=561, not a cube.a=1, b=9:C₁=1 +54 +1=56, not a cube.C₂=729 +54 +1=784=28², not a cube.a=1, b=10:C₁=1 +60 +1=62, not a cube.C₂=1000 +60 +1=1061, not a cube.So, no solutions for a=1, b>1.Similarly, for a=2, b=1:C₁=8 +12 +1=21, not a cube.C₂=1 +12 +1=14, not a cube.a=2, b=3:C₁=8 +36 +1=45, not a cube.C₂=27 +36 +1=64=4³.a=2, b=4:C₁=8 +48 +1=57, not a cube.C₂=64 +48 +1=113, not a cube.a=2, b=5:C₁=8 +60 +1=69, not a cube.C₂=125 +60 +1=186, not a cube.a=2, b=6:C₁=8 +72 +1=81=3⁴, not a cube.C₂=216 +72 +1=289=17², not a cube.a=2, b=7:C₁=8 +84 +1=93, not a cube.C₂=343 +84 +1=428, not a cube.a=2, b=8:C₁=8 +96 +1=105, not a cube.C₂=512 +96 +1=609, not a cube.a=2, b=9:C₁=8 +108 +1=117, not a cube.C₂=729 +108 +1=838, not a cube.a=2, b=10:C₁=8 +120 +1=129, not a cube.C₂=1000 +120 +1=1121, not a cube.So, no solutions for a=2, b>2.Similarly, a=3, b=1:C₁=27 +18 +1=46, not a cube.C₂=1 +18 +1=20, not a cube.a=3, b=2:C₁=27 +36 +1=64=4³C₂=8 +36 +1=45, not a cube.a=3, b=3:C₁=27 +54 +1=82, not a cube.C₂=27 +54 +1=82, not a cube.a=3, b=4:C₁=27 +72 +1=100, not a cube.C₂=64 +72 +1=137, not a cube.a=3, b=5:C₁=27 +90 +1=118, not a cube.C₂=125 +90 +1=216=6³So, here, a=3, b=5:C₁=118, not a cube.C₂=216, which is 6³.So, only C₂ is a cube.Similarly, a=3, b=6:C₁=27 +108 +1=136, not a cube.C₂=216 +108 +1=325, not a cube.a=3, b=7:C₁=27 +126 +1=154, not a cube.C₂=343 +126 +1=470, not a cube.a=3, b=8:C₁=27 +144 +1=172, not a cube.C₂=512 +144 +1=657, not a cube.a=3, b=9:C₁=27 +162 +1=190, not a cube.C₂=729 +162 +1=892, not a cube.a=3, b=10:C₁=27 +180 +1=208, not a cube.C₂=1000 +180 +1=1181, not a cube.So, only a=3, b=5 gives C₂ as a cube, but C₁ isn't.Similarly, a=4, b=3:C₁=64 +72 +1=137, not a cube.C₂=27 +72 +1=100, not a cube.a=4, b=5:C₁=64 +120 +1=185, not a cube.C₂=125 +120 +1=246, not a cube.a=4, b=6:C₁=64 +144 +1=209, not a cube.C₂=216 +144 +1=361=19², not a cube.a=4, b=7:C₁=64 +168 +1=233, not a cube.C₂=343 +168 +1=512=8³So, a=4, b=7:C₁=233, not a cube.C₂=512, which is 8³.So, only C₂ is a cube.Similarly, a=5, b=4:C₁=125 +120 +1=246, not a cube.C₂=64 +120 +1=185, not a cube.a=5, b=6:C₁=125 +180 +1=306, not a cube.C₂=216 +180 +1=397, not a cube.a=5, b=7:C₁=125 +210 +1=336, not a cube.C₂=343 +210 +1=554, not a cube.a=5, b=8:C₁=125 +240 +1=366, not a cube.C₂=512 +240 +1=753, not a cube.a=5, b=9:C₁=125 +270 +1=396, not a cube.C₂=729 +270 +1=999 +1=1000=10³So, a=5, b=9:C₁=396, not a cube.C₂=1000=10³.So, only C₂ is a cube.Similarly, a=6, b=5:C₁=216 +180 +1=397, not a cube.C₂=125 +180 +1=306, not a cube.a=6, b=7:C₁=216 +252 +1=469, not a cube.C₂=343 +252 +1=596, not a cube.a=6, b=8:C₁=216 +288 +1=505, not a cube.C₂=512 +288 +1=801, not a cube.a=6, b=9:C₁=216 +324 +1=541, not a cube.C₂=729 +324 +1=1054, not a cube.a=6, b=10:C₁=216 +360 +1=577, not a cube.C₂=1000 +360 +1=1361, not a cube.a=7, b=6:C₁=343 +252 +1=596, not a cube.C₂=216 +252 +1=469, not a cube.a=7, b=8:C₁=343 +336 +1=680, not a cube.C₂=512 +336 +1=849, not a cube.a=7, b=9:C₁=343 +378 +1=722, not a cube.C₂=729 +378 +1=1108, not a cube.a=7, b=10:C₁=343 +420 +1=764, not a cube.C₂=1000 +420 +1=1421, not a cube.a=8, b=7:C₁=512 +336 +1=849, not a cube.C₂=343 +336 +1=680, not a cube.a=8, b=9:C₁=512 +432 +1=945, not a cube.C₂=729 +432 +1=1162, not a cube.a=8, b=10:C₁=512 +480 +1=993, not a cube.C₂=1000 +480 +1=1481, not a cube.a=9, b=8:C₁=729 +432 +1=1162, not a cube.C₂=512 +432 +1=945, not a cube.a=9, b=10:C₁=729 +540 +1=1270, not a cube.C₂=1000 +540 +1=1541, not a cube.a=10, b=9:C₁=1000 +540 +1=1541, not a cube.C₂=729 +540 +1=1270, not a cube.a=10, b=10:C₁=1000 +600 +1=1601, not a cube.C₂=1000 +600 +1=1601, not a cube.So, after checking up to a=10 and b=10, the only solution I found is (1,1). Maybe that's the only solution.But to be thorough, let me check a=11, b= something.a=11:C₁=1331 +66b +1=66b +1332C₂=b³ +66b +1Looking for cubes.Let me try b=1:C₁=66 +1332=1398, not a cube.b=2:C₁=132 +1332=1464, not a cube.b=3:C₁=198 +1332=1530, not a cube.b=4:C₁=264 +1332=1596, not a cube.b=5:C₁=330 +1332=1662, not a cube.b=6:C₁=396 +1332=1728=12³. So, x=12.C₂=216 +396 +1=613, not a cube.So, no.b=7:C₁=462 +1332=1794, not a cube.b=8:C₁=528 +1332=1860, not a cube.b=9:C₁=594 +1332=1926, not a cube.b=10:C₁=660 +1332=1992, not a cube.b=11:C₁=726 +1332=2058, not a cube.b=12:C₁=792 +1332=2124, not a cube.b=13:C₁=858 +1332=2190, not a cube.b=14:C₁=924 +1332=2256, not a cube.b=15:C₁=990 +1332=2322, not a cube.b=16:C₁=1056 +1332=2388, not a cube.b=17:C₁=1122 +1332=2454, not a cube.b=18:C₁=1188 +1332=2520, not a cube.b=19:C₁=1254 +1332=2586, not a cube.b=20:C₁=1320 +1332=2652, not a cube.So, only b=6 gives C₁ as a cube, but C₂ isn't.Similarly, a=11, b=6:C₁=1728=12³C₂=216 +66*6 +1=216 +396 +1=613, not a cube.So, no.Similarly, a=12:C₁=1728 +72b +1=72b +1729C₂=b³ +72b +1Looking for cubes.b=1:C₁=72 +1729=1801, not a cube.b=2:C₁=144 +1729=1873, not a cube.b=3:C₁=216 +1729=1945, not a cube.b=4:C₁=288 +1729=2017, not a cube.b=5:C₁=360 +1729=2089, not a cube.b=6:C₁=432 +1729=2161, not a cube.b=7:C₁=504 +1729=2233, not a cube.b=8:C₁=576 +1729=2305, not a cube.b=9:C₁=648 +1729=2377, not a cube.b=10:C₁=720 +1729=2449, not a cube.b=11:C₁=792 +1729=2521, not a cube.b=12:C₁=864 +1729=2593, not a cube.b=13:C₁=936 +1729=2665, not a cube.b=14:C₁=1008 +1729=2737, not a cube.b=15:C₁=1080 +1729=2809=17², not a cube.b=16:C₁=1152 +1729=2881, not a cube.b=17:C₁=1224 +1729=2953, not a cube.b=18:C₁=1296 +1729=3025=55², not a cube.b=19:C₁=1368 +1729=3097, not a cube.b=20:C₁=1440 +1729=3169, not a cube.So, no cubes here.Similarly, a=12, b=15:C₁=2809=17², not a cube.C₂=3375 +1080 +1=4456, not a cube.So, no.I think it's safe to conclude that the only solution is (1,1). Let me just check a=0, but a must be positive, so a=0 is invalid.Therefore, the only pair of positive integers (a, b) such that both a³ +6ab +1 and b³ +6ab +1 are perfect cubes is (1,1).

answer：First, determine the number of aluminum cans collected by each person yesterday. Sarah collected 50 cans, and Lara collected 30 more than Sarah, which is 80 cans. Together, they collected 130 cans yesterday.Next, calculate the total number of cans collected today. Sarah collected 40 cans, and Lara collected 70 cans, making a total of 110 cans today.Finally, find the difference between yesterday's and today's collections by subtracting today's total from yesterday's total: 130 minus 110 equals 20 fewer cans collected today.