## 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