answer：First, I need to determine the total number of bananas collected by the family of monkeys.There are two types of piles: those with 6 piles and those with the remaining piles. For the 6 piles with 9 hands each and 14 bananas per hand:6 piles × 9 hands per pile × 14 bananas per hand = 756 bananas.For the remaining piles, there are a total of 10 piles, so 10 - 6 = 4 piles. These piles have 12 hands each and 9 bananas per hand:4 piles × 12 hands per pile × 9 bananas per hand = 432 bananas.Adding both amounts gives the total number of bananas:756 + 432 = 1,188 bananas.Finally, to find out how many bananas each monkey gets when divided equally among the 12 monkeys:1,188 bananas ÷ 12 monkeys = 99 bananas per monkey.

answer：First, I recognize that the square root of a number ( a ) is given by two expressions: ( x + 3 ) and ( 3x - 11 ). Since both expressions represent the same square root, they must be equal to each other.I set up the equation:[x + 3 = 3x - 11]Next, I solve for ( x ) by isolating the variable. Subtracting ( x ) from both sides gives:[3 = 2x - 11]Then, adding 11 to both sides:[14 = 2x]Finally, dividing both sides by 2:[x = 7]With ( x ) determined, I substitute ( x = 7 ) back into one of the expressions to find ( a ). Using ( x + 3 ):[sqrt{a} = 7 + 3 = 10]Squaring both sides to solve for ( a ):[a = 10^2 = 100]Finally, I calculate ( 2a - 1 ):[2a - 1 = 2 times 100 - 1 = 200 - 1 = 199]

answer：First, I recognize that the ladder, the wall, and the ground form a right-angled triangle. The ladder acts as the hypotenuse with a length of 8.5 meters, the height it reaches up the wall is one leg measuring 7.5 meters, and the distance from the base of the ladder to the wall is the other leg, which I need to find.Using the Pythagorean theorem, which states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides, I can set up the equation:8.5² = 7.5² + x²Calculating the squares:72.25 = 56.25 + x²Subtracting 56.25 from both sides to solve for x²:x² = 16Taking the square root of both sides gives:x = 4Therefore, the base of the ladder is 4 meters away from the wall.

answer：Okay, so I have this expression to evaluate for any natural number ( n ). It's a cube root of a fraction, where both the numerator and the denominator are sums of products. Let me write it down again to make sure I understand:[sqrt[3]{frac{1 cdot 2 cdot 4 + 2 cdot 4 cdot 8 + ldots + n cdot 2n cdot 4n}{1 cdot 3 cdot 9 + 2 cdot 6 cdot 18 + ldots + n cdot 3n cdot 9n}}]Hmm, so both the numerator and the denominator are sums of terms. Each term in the numerator is of the form ( k cdot 2k cdot 4k ), and each term in the denominator is ( k cdot 3k cdot 9k ). Let me try to generalize these terms.Starting with the numerator: Each term is ( k times 2k times 4k ). Let me compute that:( k times 2k = 2k^2 ), then ( 2k^2 times 4k = 8k^3 ). So each term in the numerator is ( 8k^3 ).Similarly, for the denominator: Each term is ( k times 3k times 9k ). Let's compute that:( k times 3k = 3k^2 ), then ( 3k^2 times 9k = 27k^3 ). So each term in the denominator is ( 27k^3 ).So, the entire numerator is the sum from ( k = 1 ) to ( n ) of ( 8k^3 ), and the denominator is the sum from ( k = 1 ) to ( n ) of ( 27k^3 ).So, the expression simplifies to:[sqrt[3]{frac{8 sum_{k=1}^{n} k^3}{27 sum_{k=1}^{n} k^3}}]Wait, both the numerator and the denominator have the same summation ( sum_{k=1}^{n} k^3 ). So, if I factor that out, the fraction becomes:[frac{8 sum_{k=1}^{n} k^3}{27 sum_{k=1}^{n} k^3} = frac{8}{27}]Because the ( sum_{k=1}^{n} k^3 ) terms cancel out. So, the expression inside the cube root is just ( frac{8}{27} ).Therefore, the cube root of ( frac{8}{27} ) is:[sqrt[3]{frac{8}{27}} = frac{sqrt[3]{8}}{sqrt[3]{27}} = frac{2}{3}]Wait, that seems straightforward. Let me double-check my steps to make sure I didn't skip anything important.1. I recognized that each term in the numerator is ( k cdot 2k cdot 4k ), which simplifies to ( 8k^3 ). That makes sense because ( k times 2k = 2k^2 ), and then ( 2k^2 times 4k = 8k^3 ).2. Similarly, each term in the denominator is ( k cdot 3k cdot 9k ), which simplifies to ( 27k^3 ). That also makes sense: ( k times 3k = 3k^2 ), and then ( 3k^2 times 9k = 27k^3 ).3. So, the numerator is ( 8 ) times the sum of ( k^3 ) from 1 to ( n ), and the denominator is ( 27 ) times the same sum. Since the sums are identical, they cancel each other out, leaving ( frac{8}{27} ).4. Taking the cube root of ( frac{8}{27} ) gives ( frac{2}{3} ), since ( 2^3 = 8 ) and ( 3^3 = 27 ).I don't see any mistakes in that reasoning. It seems that regardless of the value of ( n ), as long as it's a natural number, the expression simplifies to ( frac{2}{3} ). That's interesting because the sums in the numerator and denominator depend on ( n ), but their ratio is constant.Let me test this with a small value of ( n ), say ( n = 1 ), to see if it holds.For ( n = 1 ):Numerator: ( 1 cdot 2 cdot 4 = 8 )Denominator: ( 1 cdot 3 cdot 9 = 27 )So, the fraction is ( frac{8}{27} ), and the cube root is ( frac{2}{3} ). That works.How about ( n = 2 ):Numerator: ( 1 cdot 2 cdot 4 + 2 cdot 4 cdot 8 = 8 + 64 = 72 )Denominator: ( 1 cdot 3 cdot 9 + 2 cdot 6 cdot 18 = 27 + 216 = 243 )Fraction: ( frac{72}{243} ). Let's simplify that. Both are divisible by 9: ( 72 ÷ 9 = 8 ), ( 243 ÷ 9 = 27 ). So, ( frac{8}{27} ). Cube root is ( frac{2}{3} ). Perfect.Another test with ( n = 3 ):Numerator: ( 1 cdot 2 cdot 4 + 2 cdot 4 cdot 8 + 3 cdot 6 cdot 12 )Compute each term:1st term: ( 1 times 2 times 4 = 8 )2nd term: ( 2 times 4 times 8 = 64 )3rd term: ( 3 times 6 times 12 = 216 )Sum: ( 8 + 64 + 216 = 288 )Denominator: ( 1 cdot 3 cdot 9 + 2 cdot 6 cdot 18 + 3 cdot 9 cdot 27 )Compute each term:1st term: ( 1 times 3 times 9 = 27 )2nd term: ( 2 times 6 times 18 = 216 )3rd term: ( 3 times 9 times 27 = 729 )Sum: ( 27 + 216 + 729 = 972 )Fraction: ( frac{288}{972} ). Simplify. Both divisible by 36: ( 288 ÷ 36 = 8 ), ( 972 ÷ 36 = 27 ). So, ( frac{8}{27} ). Cube root is ( frac{2}{3} ). Yep, still holds.So, it seems consistent across different values of ( n ). Therefore, I can confidently say that the value of the expression is ( frac{2}{3} ).**Final Answer**The value of the expression is boxed{dfrac{2}{3}}.

answer：First, I need to determine the number of parents who disagree with the tuition fee increase. The total number of parents surveyed is 800, and 20% of them agree with the increase.To find the number of parents who agree, I calculate 20% of 800:20% of 800 = 0.20 × 800 = 160 parents.Since the total number of parents is 800, the number of parents who disagree is the remainder after subtracting those who agree:Number of disagreeing parents = 800 - 160 = 640 parents.

answer：Okay, so I have this problem where I'm given positive rational numbers (a, b, c) (all non-zero) and they satisfy the equation (a + b + c = abc). I need to find the value of the expression:[frac{a}{b} + frac{a}{c} + frac{b}{a} + frac{b}{c} + frac{c}{a} + frac{c}{b} - ab - bc - ca]Hmm, that looks a bit complicated, but maybe I can simplify it somehow. Let me see.First, let me write down the given equation again:[a + b + c = abc]I wonder if I can manipulate this equation to express one variable in terms of the others or find some relationship between them. Maybe I can divide both sides by (abc) to see if that helps.Dividing both sides by (abc), we get:[frac{1}{bc} + frac{1}{ac} + frac{1}{ab} = 1]Hmm, that might be useful later. Let me note that down.Now, looking at the expression I need to find:[frac{a}{b} + frac{a}{c} + frac{b}{a} + frac{b}{c} + frac{c}{a} + frac{c}{b} - ab - bc - ca]I notice that this expression has two parts: the first part is the sum of reciprocals, and the second part is the negative of the sum of products. Maybe I can group them or find a way to relate them to the given equation.Let me denote the expression as (E):[E = left( frac{a}{b} + frac{a}{c} + frac{b}{a} + frac{b}{c} + frac{c}{a} + frac{c}{b} right) - (ab + bc + ca)]I think it might help to consider each term separately or look for symmetries or substitutions.Since (a, b, c) are positive rationals, maybe I can assume some substitution or set (x = frac{1}{a}), (y = frac{1}{b}), (z = frac{1}{c}). Let me try that.Let (x = frac{1}{a}), (y = frac{1}{b}), (z = frac{1}{c}). Then, (a = frac{1}{x}), (b = frac{1}{y}), (c = frac{1}{z}).Substituting into the given equation:[frac{1}{x} + frac{1}{y} + frac{1}{z} = frac{1}{x} cdot frac{1}{y} cdot frac{1}{z}]Simplifying:[frac{1}{x} + frac{1}{y} + frac{1}{z} = frac{1}{xyz}]Multiply both sides by (xyz):[yz + xz + xy = 1]So, (xy + yz + zx = 1). That's a nice equation. Maybe this substitution will help simplify the expression (E).Let me rewrite (E) in terms of (x, y, z):First, the reciprocal terms:[frac{a}{b} = frac{frac{1}{x}}{frac{1}{y}} = frac{y}{x}]Similarly,[frac{a}{c} = frac{z}{x}, quad frac{b}{a} = frac{x}{y}, quad frac{b}{c} = frac{z}{y}, quad frac{c}{a} = frac{x}{z}, quad frac{c}{b} = frac{y}{z}]So, the first part of (E) becomes:[frac{y}{x} + frac{z}{x} + frac{x}{y} + frac{z}{y} + frac{x}{z} + frac{y}{z}]Which can be written as:[left( frac{x}{y} + frac{y}{x} right) + left( frac{x}{z} + frac{z}{x} right) + left( frac{y}{z} + frac{z}{y} right)]Each pair like (frac{x}{y} + frac{y}{x}) is equal to (frac{x^2 + y^2}{xy}). So, let me write that:[frac{x^2 + y^2}{xy} + frac{x^2 + z^2}{xz} + frac{y^2 + z^2}{yz}]Simplify each term:[frac{x^2 + y^2}{xy} = frac{x}{y} + frac{y}{x}]Similarly for the others.So, that's consistent with what I had before.Now, the second part of (E) is (- (ab + bc + ca)). Let's express that in terms of (x, y, z):[ab = frac{1}{x} cdot frac{1}{y} = frac{1}{xy}]Similarly,[bc = frac{1}{yz}, quad ca = frac{1}{zx}]So, (ab + bc + ca = frac{1}{xy} + frac{1}{yz} + frac{1}{zx})Therefore, the entire expression (E) becomes:[left( frac{x}{y} + frac{y}{x} + frac{x}{z} + frac{z}{x} + frac{y}{z} + frac{z}{y} right) - left( frac{1}{xy} + frac{1}{yz} + frac{1}{zx} right)]Hmm, that's still a bit complicated, but maybe I can find a way to relate this to the equation (xy + yz + zx = 1).Let me denote (S = x + y + z), (P = xy + yz + zx = 1), and (Q = xyz). Maybe these symmetric sums can help.Wait, but I don't know (S) or (Q). Maybe I can express the terms in (E) in terms of (S) and (P).First, let's compute the sum (frac{x}{y} + frac{y}{x} + frac{x}{z} + frac{z}{x} + frac{y}{z} + frac{z}{y}).This is equal to:[left( frac{x}{y} + frac{y}{x} right) + left( frac{x}{z} + frac{z}{x} right) + left( frac{y}{z} + frac{z}{y} right)]Each pair is (frac{x^2 + y^2}{xy}), so:[frac{x^2 + y^2}{xy} + frac{x^2 + z^2}{xz} + frac{y^2 + z^2}{yz}]Which can be written as:[frac{x^2}{xy} + frac{y^2}{xy} + frac{x^2}{xz} + frac{z^2}{xz} + frac{y^2}{yz} + frac{z^2}{yz}]Simplify each term:[frac{x}{y} + frac{y}{x} + frac{x}{z} + frac{z}{x} + frac{y}{z} + frac{z}{y}]Wait, that just brings me back to where I started. Maybe another approach.Alternatively, let me consider that:[frac{x}{y} + frac{y}{x} = frac{x^2 + y^2}{xy}]Similarly for others. So, the entire sum is:[frac{x^2 + y^2}{xy} + frac{x^2 + z^2}{xz} + frac{y^2 + z^2}{yz}]Which can be written as:[frac{x^2 + y^2 + x^2 + z^2 + y^2 + z^2}{xy + xz + yz}]Wait, no, that's not correct because each term has a different denominator. So, I can't combine them like that.Alternatively, maybe I can write each fraction as:[frac{x^2 + y^2}{xy} = frac{x}{y} + frac{y}{x}]But that doesn't help much. Maybe I can factor or find a common expression.Alternatively, let me consider that:[frac{x}{y} + frac{y}{x} = left( frac{x + y}{sqrt{xy}} right)^2 - 2]But that might complicate things more.Wait, perhaps I can consider the entire expression (E) as:[left( sum_{text{sym}} frac{x}{y} right) - left( sum frac{1}{xy} right)]Where (sum_{text{sym}} frac{x}{y}) is the symmetric sum over all permutations.But I'm not sure if that helps. Maybe I can compute (E) in terms of (x, y, z) and then use the given condition (xy + yz + zx = 1).Alternatively, maybe I can find specific values for (x, y, z) that satisfy (xy + yz + zx = 1). Since (a, b, c) are positive rationals, (x, y, z) are also positive rationals.Perhaps I can assume that (x = y = z). Let me test that.If (x = y = z), then (3x^2 = 1), so (x = frac{1}{sqrt{3}}). But that's irrational, and we need (x, y, z) to be rational. So, that's not possible.Hmm, maybe another approach. Let me consider specific values for (x, y, z) such that (xy + yz + zx = 1). For example, let me set (x = y = t), then (2t^2 + t z = 1). Maybe I can solve for (z):[z = frac{1 - 2t^2}{t}]Since (z) must be positive, (1 - 2t^2 > 0), so (t < frac{1}{sqrt{2}}). Let me choose (t = frac{1}{2}), then:[z = frac{1 - 2*(1/2)^2}{1/2} = frac{1 - 2*(1/4)}{1/2} = frac{1 - 1/2}{1/2} = frac{1/2}{1/2} = 1]So, (x = y = 1/2), (z = 1). Let me check if this satisfies (xy + yz + zx = 1):[(1/2)(1/2) + (1/2)(1) + (1/2)(1) = 1/4 + 1/2 + 1/2 = 1/4 + 1 = 5/4 neq 1]Oops, that doesn't work. Maybe another value. Let me try (t = 1/3):[z = frac{1 - 2*(1/3)^2}{1/3} = frac{1 - 2/9}{1/3} = frac{7/9}{1/3} = 7/3]Check (xy + yz + zx):[(1/3)(1/3) + (1/3)(7/3) + (1/3)(7/3) = 1/9 + 7/9 + 7/9 = (1 + 7 + 7)/9 = 15/9 = 5/3 neq 1]Still not 1. Maybe (t = 1/4):[z = frac{1 - 2*(1/4)^2}{1/4} = frac{1 - 2/16}{1/4} = frac{14/16}{1/4} = frac{7/8}{1/4} = 7/2]Check:[(1/4)(1/4) + (1/4)(7/2) + (1/4)(7/2) = 1/16 + 7/8 + 7/8 = 1/16 + 14/8 = 1/16 + 7/4 = (1 + 28)/16 = 29/16 neq 1]Hmm, not working either. Maybe this approach isn't the best. Let me think differently.Since (xy + yz + zx = 1), perhaps I can express (E) in terms of (x, y, z) and use this condition.Let me write (E) again:[E = left( frac{x}{y} + frac{y}{x} + frac{x}{z} + frac{z}{x} + frac{y}{z} + frac{z}{y} right) - left( frac{1}{xy} + frac{1}{yz} + frac{1}{zx} right)]Let me denote (A = frac{x}{y} + frac{y}{x}), (B = frac{x}{z} + frac{z}{x}), (C = frac{y}{z} + frac{z}{y}). Then, (E = A + B + C - left( frac{1}{xy} + frac{1}{yz} + frac{1}{zx} right)).But I don't see an immediate way to relate (A, B, C) to (xy + yz + zx = 1).Wait, maybe I can consider that:[frac{x}{y} + frac{y}{x} = frac{x^2 + y^2}{xy}]Similarly for others. So, (A = frac{x^2 + y^2}{xy}), (B = frac{x^2 + z^2}{xz}), (C = frac{y^2 + z^2}{yz}).So, (E = frac{x^2 + y^2}{xy} + frac{x^2 + z^2}{xz} + frac{y^2 + z^2}{yz} - left( frac{1}{xy} + frac{1}{yz} + frac{1}{zx} right))Let me combine the terms:[E = left( frac{x^2 + y^2}{xy} - frac{1}{xy} right) + left( frac{x^2 + z^2}{xz} - frac{1}{xz} right) + left( frac{y^2 + z^2}{yz} - frac{1}{yz} right)]Simplify each bracket:[frac{x^2 + y^2 - 1}{xy} + frac{x^2 + z^2 - 1}{xz} + frac{y^2 + z^2 - 1}{yz}]Hmm, that's interesting. So, each term is (frac{x^2 + y^2 - 1}{xy}), etc.But from the given condition, (xy + yz + zx = 1). Maybe I can express (x^2 + y^2) in terms of that.Wait, (x^2 + y^2 = (x + y)^2 - 2xy). But I don't know (x + y). Maybe not helpful.Alternatively, since (xy + yz + zx = 1), perhaps (x^2 + y^2 = (x + y)^2 - 2xy), but I don't see how that connects.Wait, maybe I can consider that (x^2 + y^2 = (x + y)^2 - 2xy), but without knowing (x + y), it's hard to proceed.Alternatively, perhaps I can factor the numerator (x^2 + y^2 - 1). Let me see:[x^2 + y^2 - 1 = x^2 + y^2 - (xy + yz + zx)]Since (xy + yz + zx = 1), so:[x^2 + y^2 - 1 = x^2 + y^2 - xy - yz - zx]Hmm, can I factor this?Let me write it as:[x^2 - xy + y^2 - yz - zx]Hmm, not sure. Maybe group terms:[x^2 - xy - zx + y^2 - yz][= x(x - y - z) + y(y - z)]Not particularly helpful.Wait, maybe I can think of (x^2 + y^2 - 1) as (x^2 + y^2 - (xy + yz + zx)). Let me factor:[x^2 + y^2 - xy - yz - zx = x^2 - xy + y^2 - yz - zx]Hmm, perhaps factor by grouping:[x(x - y) + y(y - z) - zx]Still not helpful.Alternatively, maybe I can write (x^2 + y^2 - 1 = (x + y)^2 - 2xy - 1), but since (xy + yz + zx = 1), we have (xy = 1 - yz - zx). So,[(x + y)^2 - 2(1 - yz - zx) - 1 = (x + y)^2 - 2 + 2yz + 2zx - 1 = (x + y)^2 + 2yz + 2zx - 3]Not sure if that helps.Maybe I'm overcomplicating this. Let me try a different approach.Since (xy + yz + zx = 1), perhaps I can express (x^2 + y^2) in terms of this.Wait, (x^2 + y^2 = (x + y)^2 - 2xy). But I don't know (x + y). Maybe if I can find (x + y + z), but I don't have that information.Alternatively, maybe I can consider that (x, y, z) are sides of a triangle or something, but that might not be relevant here.Wait, another idea: since (xy + yz + zx = 1), perhaps I can consider using substitution variables like (p = x + y + z), (q = xy + yz + zx = 1), (r = xyz). Then, maybe express (E) in terms of (p, q, r).But I don't know (p) or (r). Maybe I can find a relationship.Alternatively, perhaps I can consider that (E) can be expressed as:[E = sum_{text{sym}} frac{x}{y} - sum frac{1}{xy}]Which is:[left( frac{x}{y} + frac{y}{x} + frac{x}{z} + frac{z}{x} + frac{y}{z} + frac{z}{y} right) - left( frac{1}{xy} + frac{1}{yz} + frac{1}{zx} right)]Let me denote (u = frac{x}{y} + frac{y}{x}), (v = frac{x}{z} + frac{z}{x}), (w = frac{y}{z} + frac{z}{y}). Then, (E = u + v + w - left( frac{1}{xy} + frac{1}{yz} + frac{1}{zx} right)).But I still don't see a direct relationship.Wait, maybe I can consider that:[frac{x}{y} + frac{y}{x} = frac{x^2 + y^2}{xy}]Similarly,[frac{x}{z} + frac{z}{x} = frac{x^2 + z^2}{xz}][frac{y}{z} + frac{z}{y} = frac{y^2 + z^2}{yz}]So, (E) becomes:[frac{x^2 + y^2}{xy} + frac{x^2 + z^2}{xz} + frac{y^2 + z^2}{yz} - left( frac{1}{xy} + frac{1}{yz} + frac{1}{zx} right)]Let me factor out (frac{1}{xy}), (frac{1}{xz}), (frac{1}{yz}):[frac{x^2 + y^2 - 1}{xy} + frac{x^2 + z^2 - 1}{xz} + frac{y^2 + z^2 - 1}{yz}]Hmm, that's similar to what I had earlier.Wait, from the given condition (xy + yz + zx = 1), so (1 = xy + yz + zx). Therefore, (x^2 + y^2 - 1 = x^2 + y^2 - (xy + yz + zx)).So,[x^2 + y^2 - 1 = x^2 + y^2 - xy - yz - zx]Similarly for the other terms.So, (x^2 + y^2 - 1 = x^2 - xy + y^2 - yz - zx). Hmm, not sure.Alternatively, perhaps I can write (x^2 + y^2 - 1 = (x^2 - xy) + (y^2 - yz) - zx). Not helpful.Wait, maybe I can factor (x^2 + y^2 - xy - yz - zx). Let me try:[x^2 + y^2 - xy - yz - zx = x^2 - xy + y^2 - yz - zx][= x(x - y) + y(y - z) - zx]Not helpful. Maybe another approach.Wait, perhaps I can consider that (x, y, z) are positive rationals, so maybe I can set (x = frac{m}{n}), (y = frac{p}{q}), (z = frac{r}{s}), but that might complicate things.Alternatively, maybe I can consider that since (xy + yz + zx = 1), perhaps I can set (x = tan A), (y = tan B), (z = tan C) for some angles (A, B, C) in a triangle, because in a triangle, (tan A tan B + tan B tan C + tan C tan A = 1). But I'm not sure if that helps here.Wait, that's a good point. In a triangle, if (A + B + C = pi), then (tan A tan B + tan B tan C + tan C tan A = 1). So, maybe (x = tan A), (y = tan B), (z = tan C) for some triangle angles. But I don't know if that helps with the expression (E).Alternatively, maybe I can use substitution (x = frac{a}{b}), (y = frac{b}{c}), (z = frac{c}{a}), but that might lead to circular definitions.Wait, another idea: Let me consider that (E) can be written as:[E = left( frac{a}{b} + frac{b}{a} right) + left( frac{a}{c} + frac{c}{a} right) + left( frac{b}{c} + frac{c}{b} right) - (ab + bc + ca)]Which is:[left( frac{a}{b} + frac{b}{a} - ab right) + left( frac{a}{c} + frac{c}{a} - ac right) + left( frac{b}{c} + frac{c}{b} - bc right)]So, each group is of the form (frac{a}{b} + frac{b}{a} - ab). Let me compute this for one pair and see if there's a pattern.Let me compute (frac{a}{b} + frac{b}{a} - ab):[frac{a}{b} + frac{b}{a} - ab = frac{a^2 + b^2}{ab} - ab = frac{a^2 + b^2 - a^2 b^2}{ab}]Hmm, not sure if that helps.Wait, maybe I can factor the numerator:[a^2 + b^2 - a^2 b^2 = a^2(1 - b^2) + b^2 = a^2(1 - b^2) + b^2]Not helpful.Alternatively, perhaps I can write it as:[a^2 + b^2 - a^2 b^2 = a^2(1 - b^2) + b^2 = a^2(1 - b)(1 + b) + b^2]Still not helpful.Wait, another idea: Maybe I can use the given equation (a + b + c = abc) to express one variable in terms of the others and substitute into (E). Let me try that.From (a + b + c = abc), let me solve for (c):[c = frac{a + b}{ab - 1}]Assuming (ab neq 1). Since (a, b, c) are positive, (ab > 1), so (c) is positive.Now, let me substitute (c = frac{a + b}{ab - 1}) into (E):[E = frac{a}{b} + frac{a}{c} + frac{b}{a} + frac{b}{c} + frac{c}{a} + frac{c}{b} - ab - bc - ca]First, compute each term:1. (frac{a}{b}) is straightforward.2. (frac{a}{c} = frac{a}{frac{a + b}{ab - 1}} = frac{a(ab - 1)}{a + b})3. (frac{b}{a}) is straightforward.4. (frac{b}{c} = frac{b}{frac{a + b}{ab - 1}} = frac{b(ab - 1)}{a + b})5. (frac{c}{a} = frac{frac{a + b}{ab - 1}}{a} = frac{a + b}{a(ab - 1)})6. (frac{c}{b} = frac{frac{a + b}{ab - 1}}{b} = frac{a + b}{b(ab - 1)})Now, let's compute each of these:1. (frac{a}{b})2. (frac{a(ab - 1)}{a + b})3. (frac{b}{a})4. (frac{b(ab - 1)}{a + b})5. (frac{a + b}{a(ab - 1)})6. (frac{a + b}{b(ab - 1)})Now, let's compute the sum of these terms:[frac{a}{b} + frac{a(ab - 1)}{a + b} + frac{b}{a} + frac{b(ab - 1)}{a + b} + frac{a + b}{a(ab - 1)} + frac{a + b}{b(ab - 1)}]Let me group terms with similar denominators:Group 1: (frac{a}{b} + frac{b}{a})Group 2: (frac{a(ab - 1)}{a + b} + frac{b(ab - 1)}{a + b})Group 3: (frac{a + b}{a(ab - 1)} + frac{a + b}{b(ab - 1)})Compute each group:Group 1: (frac{a}{b} + frac{b}{a} = frac{a^2 + b^2}{ab})Group 2: (frac{a(ab - 1) + b(ab - 1)}{a + b} = frac{(a + b)(ab - 1)}{a + b} = ab - 1)Group 3: (frac{a + b}{a(ab - 1)} + frac{a + b}{b(ab - 1)} = (a + b)left( frac{1}{a(ab - 1)} + frac{1}{b(ab - 1)} right) = (a + b)left( frac{b + a}{ab(ab - 1)} right) = frac{(a + b)^2}{ab(ab - 1)})So, putting it all together, the sum of the reciprocal terms is:[frac{a^2 + b^2}{ab} + (ab - 1) + frac{(a + b)^2}{ab(ab - 1)}]Now, let's compute the negative of the product terms: (-ab - bc - ca)We already have (c = frac{a + b}{ab - 1}), so:[bc = b cdot frac{a + b}{ab - 1} = frac{b(a + b)}{ab - 1}][ca = a cdot frac{a + b}{ab - 1} = frac{a(a + b)}{ab - 1}]So, (-ab - bc - ca = -ab - frac{b(a + b)}{ab - 1} - frac{a(a + b)}{ab - 1})Combine the last two terms:[- frac{b(a + b) + a(a + b)}{ab - 1} = - frac{(a + b)(a + b)}{ab - 1} = - frac{(a + b)^2}{ab - 1}]So, the entire expression (E) is:[left( frac{a^2 + b^2}{ab} + (ab - 1) + frac{(a + b)^2}{ab(ab - 1)} right) - ab - frac{(a + b)^2}{ab - 1}]Simplify term by term:First, expand the expression:[frac{a^2 + b^2}{ab} + ab - 1 + frac{(a + b)^2}{ab(ab - 1)} - ab - frac{(a + b)^2}{ab - 1}]Combine like terms:- (frac{a^2 + b^2}{ab})- (ab - ab = 0)- (-1)- (frac{(a + b)^2}{ab(ab - 1)} - frac{(a + b)^2}{ab - 1})So, we have:[frac{a^2 + b^2}{ab} - 1 + left( frac{(a + b)^2}{ab(ab - 1)} - frac{(a + b)^2}{ab - 1} right)]Factor out (frac{(a + b)^2}{ab - 1}) from the last two terms:[frac{a^2 + b^2}{ab} - 1 + frac{(a + b)^2}{ab - 1} left( frac{1}{ab} - 1 right)]Simplify (frac{1}{ab} - 1 = frac{1 - ab}{ab}):[frac{a^2 + b^2}{ab} - 1 + frac{(a + b)^2}{ab - 1} cdot frac{1 - ab}{ab}]Note that (1 - ab = -(ab - 1)), so:[frac{a^2 + b^2}{ab} - 1 - frac{(a + b)^2 (ab - 1)}{ab(ab - 1)}]Simplify the last term:[frac{(a + b)^2 (ab - 1)}{ab(ab - 1)} = frac{(a + b)^2}{ab}]So, now the expression becomes:[frac{a^2 + b^2}{ab} - 1 - frac{(a + b)^2}{ab}]Combine the first and last terms:[frac{a^2 + b^2 - (a + b)^2}{ab} - 1]Compute (a^2 + b^2 - (a + b)^2):[a^2 + b^2 - (a^2 + 2ab + b^2) = -2ab]So, the expression becomes:[frac{-2ab}{ab} - 1 = -2 - 1 = -3]Wow, that worked out! So, the value of (E) is (-3).**Final Answer**boxed{-3}