Find all sets of four positive consecutive integers such that the sum of the cubes of the first three is the cube of the fourth.
We have
x^3 + (x + 1)^3 + (x + 2)^3 = (x + 3)^3 ... and expanding, we have
x^3 + x^3 + 3 x^2 + 3x + 1 + x^3 + 6x^2 + 12x + 8 = x^3 + 9x^2 + 27x + 27 .... simplify
3x^3 + 9x^2 + 15x + 9 = x^3 + 9x^2 + 27x + 27
2x^3 - 12x - 18 = 0
x^3 - 6x - 9 = 0 Using the Rational Zeroes Theorem, the possible roots of this are ±1, ±3 and ±9
And x = 3 is one solution
Using synthetic division. we have
3 1 0 - 6 -9
3 9 9
1 3 3 0
And the remaining polynomial is x^2 + 3x + 3 which has no "real" roots
So the integers are 3,4, 5 and 6
Check..... 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216 = 6^3
We have
x^3 + (x + 1)^3 + (x + 2)^3 = (x + 3)^3 ... and expanding, we have
x^3 + x^3 + 3 x^2 + 3x + 1 + x^3 + 6x^2 + 12x + 8 = x^3 + 9x^2 + 27x + 27 .... simplify
3x^3 + 9x^2 + 15x + 9 = x^3 + 9x^2 + 27x + 27
2x^3 - 12x - 18 = 0
x^3 - 6x - 9 = 0 Using the Rational Zeroes Theorem, the possible roots of this are ±1, ±3 and ±9
And x = 3 is one solution
Using synthetic division. we have
3 1 0 - 6 -9
3 9 9
1 3 3 0
And the remaining polynomial is x^2 + 3x + 3 which has no "real" roots
So the integers are 3,4, 5 and 6
Check..... 3^3 + 4^3 + 5^3 = 27 + 64 + 125 = 216 = 6^3