ParthKohli
Solving the diophantine equation:\[a^3 + b^3 + c^3 = 100a + 10b + c \]



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hmmm

ParthKohli
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Can you help me with that?

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find all integer solutions

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looks like we can factor something here

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did you try to expand
(a+b+c)^3

ParthKohli
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Whoo.

ParthKohli
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\[a^3+3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+b^3+3 b^2 c+3 b c^2+c^3\]

ParthKohli
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So do we have to add \(3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+3 b^2 c+3 b c^2\) to both sides?

ParthKohli
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\[(a + b + c)^3 = 3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+3 b^2 c+3 b c^2 + 100a + 10b + c \]

perl
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oh look at that

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one second

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(a+b+c)^3 = a^3+3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+b^3+3 b^2 c+3 b c^2+c^3

ParthKohli
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How would you solve it now?

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not sure

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where did you get this question, is it solvable?

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well the simple approach is , equate terms

ParthKohli
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Yes, a solution is \((1,5,3)\)

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dw:1360413980599:dw

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so if a^2 = 100 , b^2 = 10 , and c^2 = 1 , you have a solution

ParthKohli
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I didn't mention that \((a,b,c)\) all must be singledigit numbers.

perl
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ok, then you can go through all the cases , 09

ParthKohli
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I want a purely mathematical solution :

ParthKohli
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Not by guesses...