The consecutive numbers would be : x and (x+1).
We have the following equation:
x³+(x+1)³=1241
we solve this equation:
x³+(x+1)³=1241
x³+x³+3x²+3x+1=1241
2x³+3x²+3x+1-1241=0
2x³+3x²+3x-1240=0
we have to solve this expression (2x³+3x²+3x-1240=0) by Ruffini´s method.
1241=2³*5*31
The possible solution could be 2³=8
2 3 3 - 1240
8 16 152 1240
------------------------------------------------------------------
2 19 155 0
x=8
(x+1)=8+1=9
The numbers whose cubes add up to 1241 are 8 and 9.
we can check it out our answer:
8³+9³=512+729=1241
