find all integers n for which n^3=(n-1)^3+(n-2)^3+(n-3)^3
I don't think that the cubic formula will help, but I'm not sure
+23246 Guest's answer is correct.
If you wish to do this problem algetraicly:
(n - 1)3 = n3 - 3n2 + 3n - 1
(n - 2)3 = n3 - 6n2 + 12n - 8
(n - 3)3 = n3 - 9n2 + 27n - 27
Therefore:
(n - 1)3 + (n - 2)3 + (n - 3)3 = [ n3 - 3n2 + 3n - 1 ] + [ n3 - 6n2 + 12n - 8 ] + [ n3 - 9n2 + 27n - 27 ]
= 3n3 - 18n2 + 42n - 36
Since n3 = (n - 1)3 + (n - 2)3 + (n - 3)3,
n3 = 3n3 - 18n2 + 42n - 36
0 = 2n3 - 18n2 + 42n - 36
0 = n3 - 9n2 + 21n - 18
Try various factors of -18 to find an integer divisor of n3 - 9n2 + 21n - 18.
Sooner, or later, you will get to n = 6.
Since n = 6 is a divisor, n3 - 9n2 + 21n - 18 = (n - 6)(n2 - 3n + 3)
Since n2 - 3n + 3 cannot be factored using only integers, the only answer is: n = 6.
