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13 December, 15:10

Find three consecutive numbers whose cubes have a sum of 2241

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  1. 13 December, 15:15
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    Let "x" be the first integer:

    Second integer would be x+1,

    Third integer would be x+2,

    x³ + (x+1) ³ + (x+2) ³ = 2241

    Using the formula;

    (a+b) ³ = a³ + 3a²b + 3ab² + b³

    x³ + (x+1) ³ + (x+2) ³ = 2241

    x³ + x³ + 3x² + 3x + 1³ + x³ + 6x² + 12x + 2³ = 2241

    3x³ + 9x² + 15x + 9 = 2241

    dividing the whole equation by 3;

    x³ + 3x² + 5x + 3 = 747

    x³ + 3x² + 5x = 747 - 3

    x³ + 3x² + 5x = 744

    By graphing this equation, we can see that there is only 1 real root,

    that is:

    x = 8

    Hence,

    x³ = 512 (x+1) ³ = 729 (x+2) ³ = 1000
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