Equation: a^3 + b^3 – c^3 = a + b – c

 
 
Find solutions of

a^3 \; + \; b^3 \; - \; c^3 \; = \; a \; + \; b \; - \; c

 
 

Note that

Case (1) :     a^3 \; + \; b^3 \; + \; c^3 \; = \; a \; + \; b \; + \; c

a^3 \; + \; b^3 \; + \; c^3 \; = \; (a+b+c)^3 \; - \; 3 \,(a+b) \,(a+c) \,(b+c)

(a+b+c-1) \,(a+b+c) \,(a+b+c+1) \; = \; 3 \,(a+b) \,(a+c) \,(b+c)

 

Case (2) :     a^3 \; + \; b^3 \; - \; c^3 \; = \; a \; + \; b \; - \; c

a^3 \; + \; b^3 \; - \; c^3 \; = \; (a+b-c)^3 \; - \; 3 \,(a+b) \,(a-c) \,(b-c)

(a+b-c-1) \,(a+b-c) \,(a+b-c+1) \; = \; 3 \,(a+b) \,(a-c) \,(b-c)

 

                      ……………………………………………..          
 
 
The first few solutions are :

1^3 \; + \; 3^3 \; - \; 3^3 \; = \; 1 \; = \; 1 \; + \; 3 \; - \; 3

101^3 \; + \; 291^3 \; - \; 295^3 \; = \; 97 \; = \; 101 \; + \; 291 \; - \; 295

9897^3 \; + \; 28515^3 \; - \; 28907^3 \; = \; 9505 \; = \; 9897 \; + \; 28515 \; - \; 28907

969805^3 \; + \; 2794179^3 \; - \; 2832591^3
= \; 969805 \; + \; 2794179 \; - \; 2832591
= \; 931393

 
 

find the next solution.

determine the recurrence relation

 
 

 
 
 
 

 
 
 
 
 
 
 
 
 
 
 
 

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About benvitalis

math grad - Interest: Number theory
This entry was posted in Number Puzzles and tagged . Bookmark the permalink.

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