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Suppose the middle one is 3x, then the other two are 3x-3 and 3x+3.
Their product is
(3x-3)*3x*(3x+3) = 3x*(3x-3)*(3x+3) = 27x(x-1)*(x+1) = 27x*(x2-1)
So 27x*(x2-1) = 648
x*(x2-1) = 648/27 = 24
ie x3 - x - 24 = 0
or x3 - 3x2 + 3x2 - 9x + 8x - 24 = 0
ie (x - 3)(x2 + 3x + 8) = 0
The only integer solution is x = 3
So the three numbers are 3*(3-1), 3*3 and 3*(3+1) = 6, 9 and 12.
What else can I help you with?
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