Find three consecutive numbers whose cubes have a sum of 2241

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