A follow - up to 'Good night!... again."
Calculate the value of the expression above for distinct real values of , , and .
The answer is 19.
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1 solution
X = cyc ∑ ( a − b ) ( a − c ) ( 6 − 2 a + 3 b ) ( 6 − 2 a + 3 c ) = cyc ∑ ( x − y ) ( x − z ) ( 3 y − 2 x ) ( 3 z − 2 x ) = cyc ∑ ( x − y ) ( y − z ) ( z − x ) ( 9 y z − 6 x y − 6 z x + 4 x 2 ) ( z − y ) = cyc ∑ x y 2 − x 2 y + y z 2 − y 2 z + z x 2 − z 2 x 9 ( y z 2 − y 2 z ) + 6 ( x y 2 − z 2 x ) + 4 ( z x 2 − x 2 y ) = 1 9 × x y 2 − x 2 y + y z 2 − y 2 z + z x 2 − z 2 x x y 2 − x 2 y + y z 2 − y 2 z + z x 2 − z 2 x = 1 9 Let x = a + 6 , y = b + 6 , z = c + 6