Question 1155188: Find three consecutive integers whose product is 161 larger than the cube of the smallest integer Found 2 solutions by ewatrrr, MathTherapy:Answer by ewatrrr(24785) (Show Source):
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Find three consecutive integers whose product is 161 larger than the cube of the smallest integer
Let the first integer be F
Then other 2 are: F + 1, and F + 2
We then get: ----- Substituting 23F - 21F for 2F
F(3F + 23) - 7(3F + 23) = 0
F - 7 = 0 OR 3F + 23 = 0 OR 3F = - 23 (ignore)