answer：To determine the limiting sum ( s ) of the geometric series ( 4 - frac{8}{3} + frac{16}{9} - dots ), I first identify the first term and the common ratio of the series.The first term ( a ) is clearly ( 4 ).Next, I calculate the common ratio ( r ) by dividing the second term by the first term:[r = frac{-frac{8}{3}}{4} = -frac{2}{3}]Since the absolute value of ( r ) is ( frac{2}{3} ), which is less than 1, the series converges.Using the formula for the sum of an infinite geometric series:[s = frac{a}{1 - r}]I substitute the values of ( a ) and ( r ):[s = frac{4}{1 - left(-frac{2}{3}right)} = frac{4}{frac{5}{3}} = frac{12}{5} = 2.4]Therefore, the limiting sum ( s ) is 2.4.

answer：First, I recognize that ( x = 3 ) is a solution to the equation ( ax + 4 = 1 ).To find the value of ( a ), I substitute ( x = 3 ) into the equation, resulting in ( 3a + 4 = 1 ).Next, I isolate ( a ) by subtracting 4 from both sides of the equation, which gives ( 3a = -3 ).Finally, I solve for ( a ) by dividing both sides by 3, leading to ( a = -1 ).

answer：Okay, so I have this problem about a parabola. Let me try to understand it step by step. The equation given is ( C: y^2 = 2px ) where ( p > 0 ). The focus of this parabola is ( F ). There's a line passing through ( F ) at a 45-degree angle, and this line intersects the parabola at points ( A ) and ( B ). The distance between ( A ) and ( B ) is 8 units. I need to find the value of ( p ).First, I should recall some properties of parabolas. The standard form of a parabola that opens to the right is ( y^2 = 4ax ), where the focus is at ( (a, 0) ). Comparing this with the given equation ( y^2 = 2px ), it seems like ( 4a = 2p ), so ( a = frac{p}{2} ). Therefore, the focus ( F ) should be at ( left( frac{p}{2}, 0 right) ).Next, the problem mentions a line passing through ( F ) at a 45-degree angle. A 45-degree angle with respect to what? I think it's with respect to the x-axis, so the slope of the line is ( tan(45^circ) = 1 ). So, the equation of the line is ( y = x + c ). Since it passes through ( F ), which is ( left( frac{p}{2}, 0 right) ), we can substitute this point into the equation to find ( c ).Substituting ( x = frac{p}{2} ) and ( y = 0 ) into ( y = x + c ), we get:[ 0 = frac{p}{2} + c ]So, ( c = -frac{p}{2} ). Therefore, the equation of the line is ( y = x - frac{p}{2} ).Now, this line intersects the parabola ( y^2 = 2px ) at points ( A ) and ( B ). To find these points, I need to solve the system of equations:1. ( y^2 = 2px )2. ( y = x - frac{p}{2} )I can substitute equation 2 into equation 1 to find the x-coordinates of ( A ) and ( B ). Let's do that.Substituting ( y = x - frac{p}{2} ) into ( y^2 = 2px ):[ left( x - frac{p}{2} right)^2 = 2px ]Expanding the left side:[ x^2 - p x + frac{p^2}{4} = 2px ]Bring all terms to one side:[ x^2 - p x + frac{p^2}{4} - 2px = 0 ]Combine like terms:[ x^2 - 3p x + frac{p^2}{4} = 0 ]So, we have a quadratic equation in terms of ( x ):[ x^2 - 3p x + frac{p^2}{4} = 0 ]Let me denote this as ( ax^2 + bx + c = 0 ), where:- ( a = 1 )- ( b = -3p )- ( c = frac{p^2}{4} )To find the solutions for ( x ), I can use the quadratic formula:[ x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ]Plugging in the values:[ x = frac{3p pm sqrt{( -3p )^2 - 4 cdot 1 cdot frac{p^2}{4}}}{2 cdot 1} ]Simplify inside the square root:[ sqrt{9p^2 - p^2} = sqrt{8p^2} = 2psqrt{2} ]So, the solutions are:[ x = frac{3p pm 2psqrt{2}}{2} ]Factor out ( p ):[ x = p cdot frac{3 pm 2sqrt{2}}{2} ]So, the two x-coordinates are:[ x_1 = p cdot frac{3 + 2sqrt{2}}{2} ][ x_2 = p cdot frac{3 - 2sqrt{2}}{2} ]Now, let's find the corresponding y-coordinates using equation 2: ( y = x - frac{p}{2} ).For ( x_1 ):[ y_1 = p cdot frac{3 + 2sqrt{2}}{2} - frac{p}{2} = p cdot left( frac{3 + 2sqrt{2} - 1}{2} right) = p cdot frac{2 + 2sqrt{2}}{2} = p (1 + sqrt{2}) ]For ( x_2 ):[ y_2 = p cdot frac{3 - 2sqrt{2}}{2} - frac{p}{2} = p cdot left( frac{3 - 2sqrt{2} - 1}{2} right) = p cdot frac{2 - 2sqrt{2}}{2} = p (1 - sqrt{2}) ]So, the coordinates of points ( A ) and ( B ) are:- ( A left( p cdot frac{3 + 2sqrt{2}}{2}, p (1 + sqrt{2}) right) )- ( B left( p cdot frac{3 - 2sqrt{2}}{2}, p (1 - sqrt{2}) right) )Now, I need to find the distance between ( A ) and ( B ), which is given as 8. Let's compute the distance using the distance formula:[ |AB| = sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} ]First, compute ( x_1 - x_2 ):[ x_1 - x_2 = p cdot frac{3 + 2sqrt{2}}{2} - p cdot frac{3 - 2sqrt{2}}{2} = p cdot frac{3 + 2sqrt{2} - 3 + 2sqrt{2}}{2} = p cdot frac{4sqrt{2}}{2} = 2psqrt{2} ]Next, compute ( y_1 - y_2 ):[ y_1 - y_2 = p (1 + sqrt{2}) - p (1 - sqrt{2}) = p (1 + sqrt{2} - 1 + sqrt{2}) = p (2sqrt{2}) ]So, both differences ( x_1 - x_2 ) and ( y_1 - y_2 ) are equal to ( 2psqrt{2} ).Now, plug these into the distance formula:[ |AB| = sqrt{(2psqrt{2})^2 + (2psqrt{2})^2} ]Compute each term:[ (2psqrt{2})^2 = 4p^2 cdot 2 = 8p^2 ]So, we have:[ |AB| = sqrt{8p^2 + 8p^2} = sqrt{16p^2} = 4p ]But the problem states that ( |AB| = 8 ). Therefore:[ 4p = 8 ]Solving for ( p ):[ p = frac{8}{4} = 2 ]So, the value of ( p ) is 2.Wait, let me double-check my steps to make sure I didn't make a mistake.1. I found the focus correctly: ( F left( frac{p}{2}, 0 right) ).2. The line through ( F ) at 45 degrees has a slope of 1, so equation ( y = x - frac{p}{2} ). That seems right.3. Substituted into the parabola equation, expanded, and got quadratic in x. Let me check that again: - ( (x - p/2)^2 = 2px ) - ( x^2 - p x + p^2/4 = 2 p x ) - ( x^2 - 3 p x + p^2/4 = 0 ). Yes, that's correct.4. Applied quadratic formula: - Discriminant: ( 9p^2 - p^2 = 8p^2 ). Square root is ( 2psqrt{2} ). So, x solutions are ( [3p ± 2p√2]/2 ). Correct.5. Found y coordinates by plugging back into ( y = x - p/2 ). Calculations look correct.6. Calculated differences ( x1 - x2 = 2p√2 ) and ( y1 - y2 = 2p√2 ). Then distance is sqrt( (2p√2)^2 + (2p√2)^2 ) = sqrt(8p² + 8p²) = sqrt(16p²) = 4p. So, 4p = 8 => p=2. That seems consistent.I think all steps are correct. So, the value of ( p ) is 2.**Final Answer**The value of ( p ) is boxed{2}.

answer：Alright, so I've got this problem here about loggers harvesting wood. Let me try to parse it step by step.First, the team was supposed to harvest 216 cubic meters of wood over several days. That's the total amount. Now, for the first three days, they met the daily target. So, if the daily target is, say, 'x' cubic meters per day, then for the first three days, they harvested 3x cubic meters.Then, after those three days, they started harvesting an additional 8 cubic meters each day above the daily target. So, from day four onwards, they were harvesting x + 8 cubic meters each day.Because of this increased harvesting, they ended up finishing one day ahead of schedule. So, instead of taking the originally planned number of days, they finished a day early. And the total amount they harvested was 232 cubic meters.Wait, hold on. The total amount is 232, which is more than the original 216. So, they not only finished early but also exceeded the total amount. Interesting.So, let's break this down. Let me denote:- Let the original number of days planned be 'd'.- The daily target is 'x' cubic meters.- So, the total planned amount is d * x = 216.But they ended up harvesting 232 cubic meters in (d - 1) days. Because they finished one day early.So, the total harvested is 232 = 3x + (d - 4)(x + 8). Wait, why (d - 4)? Because the first three days are at x, and then from day four to day (d - 1), which is (d - 1 - 3) = (d - 4) days, they harvested x + 8 each day.Wait, let me check that. If the total number of days they actually worked is (d - 1), and the first three days are at x, then the remaining days are (d - 1 - 3) = (d - 4) days. So, yeah, that makes sense.So, putting it all together:Total harvested = 3x + (d - 4)(x + 8) = 232.And we also know that the original plan was d * x = 216.So, now we have two equations:1. d * x = 2162. 3x + (d - 4)(x + 8) = 232We can solve these two equations to find x.Let me write them again:1. d * x = 2162. 3x + (d - 4)(x + 8) = 232First, let's express d from the first equation: d = 216 / x.Then, substitute d into the second equation.So, equation 2 becomes:3x + ( (216 / x) - 4 )(x + 8) = 232Let me compute this step by step.First, expand the second term:( (216 / x) - 4 )(x + 8) = (216 / x)(x + 8) - 4(x + 8)Compute each part:(216 / x)(x + 8) = 216 + (216 * 8)/x = 216 + 1728 / xAnd -4(x + 8) = -4x - 32So, putting it together:3x + [216 + 1728 / x - 4x - 32] = 232Simplify inside the brackets:216 - 32 = 184So, 3x + [184 + 1728 / x - 4x] = 232Combine like terms:3x - 4x = -xSo, now we have:-x + 184 + 1728 / x = 232Bring all terms to one side:-x + 184 + 1728 / x - 232 = 0Simplify constants:184 - 232 = -48So, -x - 48 + 1728 / x = 0Multiply both sides by x to eliminate the denominator:-x^2 - 48x + 1728 = 0Multiply both sides by -1 to make the quadratic coefficient positive:x^2 + 48x - 1728 = 0So, now we have a quadratic equation:x^2 + 48x - 1728 = 0Let me try to solve this quadratic equation. I can use the quadratic formula:x = [ -b ± sqrt(b^2 - 4ac) ] / 2aWhere a = 1, b = 48, c = -1728Compute discriminant:b^2 - 4ac = 48^2 - 4*1*(-1728) = 2304 + 6912 = 9216sqrt(9216) = 96So, x = [ -48 ± 96 ] / 2We have two solutions:x = ( -48 + 96 ) / 2 = 48 / 2 = 24x = ( -48 - 96 ) / 2 = -144 / 2 = -72Since x represents cubic meters per day, it can't be negative. So, x = 24.So, the daily target was 24 cubic meters.Let me verify this.If the daily target is 24, then the original number of days was d = 216 / 24 = 9 days.So, they were supposed to work 9 days, harvesting 24 m³ each day.But they harvested 3 days at 24 m³, so 3 * 24 = 72 m³.Then, from day 4 to day 8 (since they finished one day early, day 8 instead of day 9), that's 5 days.Each day, they harvested 24 + 8 = 32 m³.So, 5 days * 32 m³/day = 160 m³.Total harvested: 72 + 160 = 232 m³, which matches the given total.So, that checks out.Therefore, the team was supposed to harvest 24 cubic meters per day.**Final Answer**The team was supposed to harvest boxed{24} cubic meters of wood per day.

answer：Okay, so I came across this interesting number theory problem: Does there exist infinitely many positive integers ( k ) such that ( k cdot 2^{n} + 1 ) is composite for every positive integer ( n )? Hmm, that's a bit tricky, but let me try to break it down.First, let me understand what the problem is asking. We need to find if there are infinitely many ( k ) values where, no matter what positive integer ( n ) you choose, the expression ( k cdot 2^{n} + 1 ) always results in a composite number. That means for each such ( k ), every number in the sequence ( k cdot 2 + 1, k cdot 4 + 1, k cdot 8 + 1, ldots ) is composite. So, none of these numbers can be prime.I remember hearing about something called "Sierpiński numbers." Let me recall. A Sierpiński number is an odd positive integer ( k ) such that ( k cdot 2^{n} + 1 ) is composite for all positive integers ( n ). So, this problem is essentially asking if there are infinitely many Sierpiński numbers.Wait, but the problem doesn't specify that ( k ) has to be odd. It just says positive integers ( k ). So, maybe the question is about both even and odd ( k ). But I think Sierpiński numbers are specifically odd. Maybe there's a similar concept for even numbers?Let me check. I think there's something called "Riesel numbers," which are similar but for the form ( k cdot 2^{n} - 1 ). So, Riesel numbers are odd integers ( k ) such that ( k cdot 2^{n} - 1 ) is composite for all ( n ). So, similar concept but subtracting 1 instead of adding.But in our problem, it's adding 1, so it's more like Sierpiński numbers. So, if we're talking about ( k cdot 2^{n} + 1 ), then Sierpiński numbers are the ones where this is always composite, and they have to be odd. So, for even ( k ), maybe the problem is different.Wait, the problem allows ( k ) to be any positive integer, so both even and odd. So, perhaps we can have both even and odd ( k ) satisfying the condition. But I think Sierpiński numbers are specifically odd, so maybe the even ( k ) can be handled differently.Let me think about even ( k ). Suppose ( k ) is even, say ( k = 2m ). Then ( k cdot 2^{n} + 1 = 2m cdot 2^{n} + 1 = m cdot 2^{n+1} + 1 ). So, that's similar to the original form but shifted by one exponent. So, if ( m ) is a Sierpiński number, then ( k = 2m ) would also satisfy the condition because ( m cdot 2^{n+1} + 1 ) is composite for all ( n geq 1 ), which translates to ( k cdot 2^{n} + 1 ) being composite for all ( n geq 1 ).But wait, does that mean if we have one Sierpiński number, we can generate infinitely many by multiplying by powers of 2? Because if ( k ) is Sierpiński, then ( 2k, 4k, 8k, ldots ) would also be Sierpiński numbers? Hmm, that might be the case.But I need to verify if that's true. Let's take an example. The smallest known Sierpiński number is 78557. So, if I take ( k = 78557 ), then ( 2k = 157114 ). Is 157114 also a Sierpiński number? Let's see. For ( n = 1 ), ( 157114 cdot 2 + 1 = 314229 ). Is that composite? Let me check. 314229 divided by 3 is 104743, which is an integer, so yes, it's composite. For ( n = 2 ), ( 157114 cdot 4 + 1 = 628457 ). Is that composite? Let me see. 628457 divided by 7 is approximately 89779.571, not an integer. Divided by 11? 628457 ÷ 11 is about 57132.454, not an integer. Hmm, maybe it's prime? Wait, I'm not sure. Maybe I should look it up or use a primality test.Alternatively, maybe I can think of another approach. If ( k ) is a Sierpiński number, then ( k cdot 2^{n} + 1 ) is composite for all ( n ). So, if ( k ) is even, say ( k = 2m ), then ( 2m cdot 2^{n} + 1 = m cdot 2^{n+1} + 1 ). So, if ( m ) is a Sierpiński number, then ( m cdot 2^{n+1} + 1 ) is composite for all ( n geq 1 ), which means ( k cdot 2^{n} + 1 ) is composite for all ( n geq 1 ). So, yes, if ( m ) is Sierpiński, then ( k = 2m ) is also Sierpiński. Therefore, starting from one Sierpiński number, we can generate infinitely many by multiplying by powers of 2. So, that would give us infinitely many ( k ) such that ( k cdot 2^{n} + 1 ) is composite for all ( n ).But wait, is that correct? Because if ( k ) is even, then ( k cdot 2^{n} + 1 ) is odd, right? So, it's an odd number, which could be prime or composite. But if ( k ) is a multiple of a Sierpiński number, then it's composite for all ( n ). So, if we have one Sierpiński number, we can generate infinitely many by scaling it with powers of 2.But does that mean there are infinitely many Sierpiński numbers? Because if we can generate infinitely many from one, then yes. But I think the known Sierpiński numbers are finite, but it's conjectured that there are infinitely many. Wait, actually, I think it's proven that there are infinitely many Sierpiński numbers. Let me recall.I remember that the proof involves showing that there are infinitely many ( k ) such that ( k cdot 2^{n} + 1 ) is always composite. The method involves using covering congruences, similar to how Sierpiński originally constructed his example. So, perhaps by constructing such ( k ) with appropriate congruence conditions, we can ensure that ( k cdot 2^{n} + 1 ) is divisible by some prime for each ( n ), hence composite.Let me think about covering congruences. A covering system is a set of congruences such that every integer satisfies at least one congruence in the set. So, if we can find a set of primes and exponents such that for every ( n ), ( k cdot 2^{n} + 1 ) is divisible by at least one prime in the set, then ( k ) would be a Sierpiński number.For example, in the case of 78557, it's known that for every ( n ), ( 78557 cdot 2^{n} + 1 ) is divisible by one of the primes in the set {3, 5, 7, 13, 19, 37, 73}. So, each ( n ) falls into one of these congruence classes modulo the order of 2 modulo each prime.So, if we can find such a covering set for a ( k ), then ( k ) is a Sierpiński number. Now, to show that there are infinitely many such ( k ), perhaps we can construct them by multiplying 78557 by different powers of 2, as I thought earlier, but I need to make sure that the covering congruences still hold.Wait, actually, if ( k ) is a multiple of 78557, say ( k = 78557 cdot 2^{m} ), then ( k cdot 2^{n} + 1 = 78557 cdot 2^{m + n} + 1 ). So, for each ( n ), ( m + n ) is just another exponent. Since the covering congruences for 78557 work for all exponents, then ( 78557 cdot 2^{m + n} + 1 ) will still be divisible by one of the primes in the covering set. Therefore, ( k = 78557 cdot 2^{m} ) is also a Sierpiński number for any non-negative integer ( m ). Hence, we can generate infinitely many Sierpiński numbers this way.But wait, does this hold? Let me test with ( m = 1 ). So, ( k = 78557 cdot 2 = 157114 ). For ( n = 1 ), ( 157114 cdot 2 + 1 = 314229 ). Divided by 3: 314229 ÷ 3 = 104743, which is an integer, so composite. For ( n = 2 ), ( 157114 cdot 4 + 1 = 628457 ). Let me check if this is divisible by 5: 628457 ÷ 5 = 125691.4, not integer. Divided by 7: 628457 ÷ 7 ≈ 89779.571, not integer. Divided by 13: 628457 ÷ 13 ≈ 48342.846, not integer. Hmm, maybe it's prime? Wait, but according to the covering set, it should be divisible by one of the primes. Maybe I made a mistake.Wait, no, because when ( k = 157114 ), the exponent ( n ) is being added to the exponent in ( k ). So, the covering congruences are based on ( m + n ). So, for ( k = 78557 cdot 2^{m} ), the exponent becomes ( m + n ). So, the covering set still applies because it's just shifting the exponent. So, for each ( n ), ( m + n ) will cover all residues modulo the orders of 2 modulo each prime in the covering set. Therefore, ( k cdot 2^{n} + 1 ) will still be divisible by one of the primes in the covering set, hence composite.Therefore, by scaling the known Sierpiński number 78557 by powers of 2, we can generate infinitely many Sierpiński numbers. Hence, the answer to the problem is yes, there exist infinitely many positive integers ( k ) such that ( k cdot 2^{n} + 1 ) is composite for every positive integer ( n ).But wait, I should also consider whether there are other ways to construct such ( k ) without relying on the known Sierpiński number. Maybe there are other methods or other covering sets that can be used to construct different Sierpiński numbers. But regardless, even if we can generate infinitely many from one example, that's sufficient to answer the question affirmatively.Another thought: maybe even numbers can also be Sierpiński numbers. Wait, no, because if ( k ) is even, say ( k = 2m ), then ( k cdot 2^{n} + 1 = m cdot 2^{n+1} + 1 ). So, if ( m ) is a Sierpiński number, then yes, ( k ) would also be a Sierpiński number. But if ( m ) is not a Sierpiński number, then ( k ) might not be. So, the key is that if we can find one Sierpiński number, we can generate infinitely many by scaling it with powers of 2.Therefore, since we know that 78557 is a Sierpiński number, and by multiplying it by ( 2^{m} ) for any ( m geq 0 ), we get another Sierpiński number, and since there are infinitely many such ( m ), there are infinitely many ( k ) satisfying the condition.Wait, but does this scaling method actually produce distinct Sierpiński numbers? Yes, because each scaling by a different power of 2 gives a different ( k ). So, ( 78557, 157114, 314228, ldots ) are all distinct and each is a Sierpiński number.Therefore, the answer is yes, there are infinitely many such ( k ).**Final Answer**boxed{Yes}

answer：Alright, so I have this expression to evaluate: (2000^3 - 1999 cdot 2000^2 - 1999^2 cdot 2000 + 1999^3). Hmm, that looks a bit complicated at first glance, but maybe I can simplify it somehow. Let me try to break it down step by step.First, I notice that both 2000 and 1999 are involved in all the terms. Maybe I can factor something out or find a common pattern. Let me write down the expression again:(2000^3 - 1999 cdot 2000^2 - 1999^2 cdot 2000 + 1999^3).Looking at this, I see that the first two terms have (2000^2) in them, and the last two terms have (1999^2). Maybe I can factor out (2000^2) from the first two terms and (1999^2) from the last two. Let me try that:(2000^2(2000 - 1999) - 1999^2(2000 - 1999)).Wait, that seems promising. Let me compute the expressions inside the parentheses first. (2000 - 1999) is just 1. So substituting that back in, the expression becomes:(2000^2 cdot 1 - 1999^2 cdot 1).Which simplifies to:(2000^2 - 1999^2).Oh, now this looks familiar! It's a difference of squares, right? The formula for that is (a^2 - b^2 = (a - b)(a + b)). So applying that here, where (a = 2000) and (b = 1999), we get:((2000 - 1999)(2000 + 1999)).Calculating each part, (2000 - 1999) is 1, and (2000 + 1999) is 3999. So multiplying those together:(1 times 3999 = 3999).Wait, so does that mean the entire original expression simplifies down to 3999? Let me double-check my steps to make sure I didn't make a mistake.Starting from the original expression:(2000^3 - 1999 cdot 2000^2 - 1999^2 cdot 2000 + 1999^3).I factored out (2000^2) from the first two terms:(2000^2(2000 - 1999)).Which is correct because (2000^3 = 2000^2 cdot 2000) and (1999 cdot 2000^2) is just (1999 cdot 2000^2), so factoring that out leaves (2000 - 1999).Similarly, for the last two terms, I factored out (-1999^2):(-1999^2(2000 - 1999)).Wait, hold on. Let me check that again. The original expression is:(-1999^2 cdot 2000 + 1999^3).So factoring out (1999^2), it becomes (1999^2(-2000 + 1999)), which is (1999^2(-1)). So actually, the expression after factoring should be:(2000^2(2000 - 1999) - 1999^2(2000 - 1999)).Which is (2000^2(1) - 1999^2(1)), so (2000^2 - 1999^2). So that part is correct.Then, applying the difference of squares formula:((2000 - 1999)(2000 + 1999) = 1 times 3999 = 3999).So it seems my steps are correct. Therefore, the value of the expression is 3999.But just to be thorough, let me compute each term individually and then add them up to see if I get the same result. Maybe that will help confirm.First, compute (2000^3):(2000^3 = 2000 times 2000 times 2000 = 8,000,000,000).Next, compute (1999 cdot 2000^2):(2000^2 = 4,000,000), so (1999 times 4,000,000 = 7,996,000,000).Then, compute (1999^2 cdot 2000):First, (1999^2 = (2000 - 1)^2 = 2000^2 - 2 times 2000 times 1 + 1^2 = 4,000,000 - 4,000 + 1 = 3,996,001).Then, multiply that by 2000: (3,996,001 times 2000 = 7,992,002,000).Lastly, compute (1999^3):(1999^3 = 1999 times 1999 times 1999). Hmm, that's a bit more involved. Let me compute (1999 times 1999) first, which we already did as 3,996,001. Then multiply that by 1999:(3,996,001 times 1999).Let me compute this step by step. First, note that (3,996,001 times 2000 = 7,992,002,000). Then subtract (3,996,001) to get (3,996,001 times 1999):(7,992,002,000 - 3,996,001 = 7,992,002,000 - 3,996,001 = 7,988,006,000 - 0,000,001 = 7,988,005,999).Wait, let me verify that subtraction:(7,992,002,000 - 3,996,001).Subtracting 3,996,001 from 7,992,002,000:First, subtract 3,000,000: 7,992,002,000 - 3,000,000 = 7,989,002,000.Then subtract 996,001: 7,989,002,000 - 996,001 = 7,988,006,000 - 0,000,001 = 7,988,005,999.Yes, that seems correct.So now, putting it all together:(2000^3 - 1999 cdot 2000^2 - 1999^2 cdot 2000 + 1999^3)= (8,000,000,000 - 7,996,000,000 - 7,992,002,000 + 7,988,005,999).Let me compute each subtraction and addition step by step.First, (8,000,000,000 - 7,996,000,000 = 4,000,000).Then, subtract (7,992,002,000) from that result: (4,000,000 - 7,992,002,000). Wait, that would be negative. Let me write it as:(4,000,000 - 7,992,002,000 = -7,988,002,000).Then, add (7,988,005,999) to that: (-7,988,002,000 + 7,988,005,999).Calculating that: The difference between 7,988,005,999 and 7,988,002,000 is 3,999. So since it's negative 7,988,002,000 plus positive 7,988,005,999, that's 3,999.So, the total is 3,999.Wait, that's the same result as before! So that confirms it. The value is indeed 3,999.Just to recap, the original expression simplifies via factoring to a difference of squares, which then simplifies to 3,999. Computing each term individually and then combining them also leads to the same result. So I'm confident that 3,999 is the correct answer.I think I've covered all the steps and double-checked my work, so I don't see any mistakes in my reasoning. It's interesting how factoring can simplify such a complex-looking expression into something manageable. I should remember to look for factoring opportunities in the future when dealing with polynomials or expressions with multiple terms.Another thing I noticed is that 2000 and 1999 are consecutive integers, which might have hinted at using the difference of squares or some similar identity. It's useful to recognize such patterns because they can save a lot of computation time, especially with large numbers like these.In summary, by factoring and applying the difference of squares formula, I was able to simplify the expression and find that its value is 3,999. Verifying by expanding each term also confirmed this result, so I'm pretty sure that's the correct answer.**Final Answer**boxed{3999}