2 0 1 6 x + 2 0 1 6 − x = 3
2 0 1 6 x − 2 0 1 6 − x 2 0 1 6 6 x − 2 0 1 6 − 6 x = ?
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What a hard problem! I thought too much about it!
From the 5th line, I think it would be clearer to factorise the a 2 + a − 2 + 1 and a 2 + a − 2 − 1 in terms of perfect squares. Aside from that, nice answer!
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Let a = 2 0 1 6 x . Then 2 0 1 6 x + 2 0 1 6 − x = a + a − 1 = 3 and we have:
X = 2 0 1 6 x − 2 0 1 6 − x 2 0 1 6 6 x − 2 0 1 6 − 6 x = a − a − 1 a 6 − a − 6 = a − a − 1 ( a 3 − a − 3 ) ( a 3 + a − 3 ) = a − a − 1 ( a − a − 1 ) ( a 2 + a − 2 + 1 ) ( a + a − 1 ) ( a 2 + a − 2 − 1 ) = ( a 2 + a − 2 + 1 ) ( 3 ) ( a 2 + a − 2 − 1 ) = 3 ( ( a + a − 1 ) 2 − 1 ) ( ( a + a − 1 ) 2 − 3 ) = 3 ( 9 − 1 ) ( 9 − 3 ) = 3 ( 8 ) ( 6 ) = 1 4 4 = 1 2