MHB Solving $x^3+y^3+z^3=(x+y+z)^2$ with Positive Integers

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Find all solutions in positive integers $z<y<x$ to the equation $x^3+y^3+z^3=(x+y+z)^2$.
 
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anemone said:
Find all solutions in positive integers $z<y<x$ to the equation $x^3+y^3+z^3=(x+y+z)^2$.

first let us find upper bound for x

even if x = y = z we get $3x^3= (3x)^2 = 9x^2$ or x = 3

for lower bound as z < y < x so minimum value of x = 3

so we get x = 3, y = 2 and z = 1 is the only case and check that it satisfies the condition

as $3^3+2^3+1^3 = (3+2+1)^3 = 36$ so this is the solution

hence

$(x,y,z) = (3,2,1)$
 
Very well done, Kali! Thanks for participating too!:)
 
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