Let "x" be the first integer:
Second integer would be x+1,
Third integer would be x+2,
<h3>x³ + (x+1)³ + (x+2)³ = 2241</h3>
Using the formula;
<h3>(a+b)³ = a³ + 3a²b + 3ab² + b³ </h3>
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
<h2>x³ + 3x² + 5x = 744</h2>
By graphing this equation, we can see that there is only 1 real root,
that is:
x = 8
Hence,
<h2>
x³ = 512</h2><h2>
(x+1)³ = 729</h2><h2>
(x+2)³ = 1000</h2>
