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Let the two consecutive numbers be ( n ) and ( n+1 ). The difference of their cubes can be expressed as ( (n+1)^3 - n^3 = 3n^2 + 3n + 1 ). Setting this equal to 631 gives the equation ( 3n^2 + 3n + 1 = 631 ). Solving for ( n ) leads to ( 3n^2 + 3n - 630 = 0 ), which simplifies to ( n^2 + n - 210 = 0 ). Factoring or using the quadratic formula, we find ( n = 14 ) or ( n = -15 ). Therefore, the consecutive numbers are 14 and 15.

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