Saturday, June 22, 2013

please help me solve this: the sum of the cubes of two consecutive numbers is one more than a perfect cube. what are these two numbers?


 
which gives us...
%0D%0A%282n%2B1%29%28n%5E2%2Bn%2B1%29%0D%0A 
This number is to be 1 more than a perfect cube i.e. 

%0D%0Am%5E3=2n%5E3%2B3n%5E2%2B3n=n%282n%5E3%2B3n%2B3%29%0D%0A 
We only have one equation so are algebra methods end here. 


By trial and error n=1 

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