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5 April, 12:54

Find three consecutive integers whose products is 85 larger than the cube of the smallest integer

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  1. 5 April, 15:46
    0
    Let the numbers be x, x + 1 and x + 2.

    From the information given:-

    x (x + 1) (x + 2) = x^3 + 85

    Solving for x:-

    x (x^2 + 3x + 2) = x^3 + 85

    x^3 + 3x^2 + 2x = x^3 + 85

    3x^2 + 2x - 85 = 0

    3x^2 - 15x + 17x - 85 = 0

    3x (x - 5) + 17 (x - 5) = 0

    (3x + 17) (x - 5) = 0

    x = - 17/3, 5

    We take x = 5 because x is an integer.

    Answer: - the 3 integers are 5, 6 and 7.
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