THE EQUATION $X^3 + Y^3 = Z^3$ HAS NO INTEGER SOLUTIONS

     
If displaystylex+y+z=6 & displaystylex^2+y^2+z^2=14 & displaystylex^3+y^3+z^3=36 then what are displaystylex,y,z ?
displaystyleleftlbracex,y,z ight brace=leftlbrace1,2,3 ight brace Explanation: displaystyle1+2+3=6displaystyle1^2+2^2+3^2=1+4+9=14 ...

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The answer depends on what sorts of numbers you are looking for. Assuming you're looking for positive integers, at least one of the three numbers (let's say z) has khổng lồ be at least 5, since 4^3 + 4^3 + 4^3 = 192
Take your second equation và solve for Y in terms of X. Then, plug this into your first equation, & into the 3rd equation to get two equations that have X and Z as variables. Then in the first ...
(Spoiler: This answer contains the word “SEX”.) It depends. If you want three real numbers x,y,z satisfying x^3+y^3+z^3=33 then there are infinitely many, & they are very easy lớn find. If ...

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With some forethought, you can bring it down to one case to kiểm tra directly, which turns out to lớn be a solution. A previous version of this answer examined the equation thủ thuật 7, which led to lớn 16 cases ...

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https://math.stackexchange.com/questions/1762614/what-is-the-way-to-solve-the-following-system-of-equations
Assuming the last equation is meant khổng lồ be x^3 + y^3 + z^3 = 64, we can consider squaring (x + y + z): 16 = (x+y+z)^2 = (x^2 + y^2 + z^2) + 2(xy + yz + xz) = 24 + 2(xy + yz + xz) và so (xy + yz + xz) = -4 ...
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left< eginarray l l 2 & 3 \ 5 và 4 endarray ight> left< eginarray l l l 2 và 0 & 3 \ -1 & 1 và 5 endarray ight>
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