ParthKohli
  • ParthKohli
Solving the diophantine equation:\[a^3 + b^3 + c^3 = 100a + 10b + c \]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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perl
  • perl
hmmm
ParthKohli
  • ParthKohli
Can you help me with that?
perl
  • perl
find all integer solutions

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More answers

perl
  • perl
looks like we can factor something here
perl
  • perl
did you try to expand (a+b+c)^3
ParthKohli
  • ParthKohli
Whoo.
ParthKohli
  • ParthKohli
\[a^3+3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+b^3+3 b^2 c+3 b c^2+c^3\]
ParthKohli
  • ParthKohli
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
  • ParthKohli
\[(a + b + c)^3 = 3 a^2 b+3 a^2 c+3 a b^2+6 a b c+3 a c^2+3 b^2 c+3 b c^2 + 100a + 10b + c \]
perl
  • perl
oh look at that
perl
  • perl
one second
perl
  • perl
(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
  • ParthKohli
How would you solve it now?
perl
  • perl
not sure
perl
  • perl
where did you get this question, is it solvable?
perl
  • perl
well the simple approach is , equate terms
ParthKohli
  • ParthKohli
Yes, a solution is \((1,5,3)\)
perl
  • perl
|dw:1360413980599:dw|
perl
  • perl
so if a^2 = 100 , b^2 = 10 , and c^2 = 1 , you have a solution
ParthKohli
  • ParthKohli
I didn't mention that \((a,b,c)\) all must be single-digit numbers.
perl
  • perl
ok, then you can go through all the cases , 0-9
ParthKohli
  • ParthKohli
I want a purely mathematical solution :-|
ParthKohli
  • ParthKohli
Not by guesses...

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