#2**0 **

A formula for (a + b)^{3} = a^{3} + 3·a^{2}·b + a·b^{2} + b^{3}.

[You can check this formula by multiplication of (a+b)·(a + b)·(a + b).]

For (n - 1)^{3} by using the above formula ---> a = n and b = -1:

(a + b)^{3} = a^{3} + 3·a^{2}·b + 3·a·b^{2} + b^{3}.

(n - 1)^{3} = (n)^{3} + 3·(n)^{2}·(-1) + 3·(n)·(-1)^{2} + (-1)^{3}.

= n^{3} - 3n^{2} + 3n - 1

For (n - 2)^{3} by using the above formula ---> a = n and b = -2:

(a + b)^{3} = a^{3} + 3·a^{2}·b + 3·a·b^{2} + b^{3}.

(n - 1)^{3} = (n)^{3} + 3·(n)^{2}·(-2) + 3·(n)·(-2)^{2} + (-2)^{3}.

= n^{3} - 6n^{2} +12n - 8

For (n - 3)^{3} by using the above formula ---> a = n and b = -3:

(a + b)^{3} = a^{3} + 3·a^{2}·b + 3·a·b^{2} + b^{3}.

(n - 3)^{3} = (n)^{3} + 3·(n)^{2}·(-3) + 3·(n)·(-3)^{2} + (-3)^{3}.

= n^{3} - 9n^{2} +27n - 27

Putting all this together:

n^{3} = (n - 1)^{3} + (n - 2)^{3} + (n - 3)^{3}

n^{3} = ( n^{3} - 3n^{2} + 3n - 1 ) + ( n^{3} - 6n^{2} +12n - 8 ) + ( n^{3} - 9n^{2} +27n - 27 )

n^{3} = 3n^{3} -18n^{2} + 42n - 36

0 = 2n^{3} - 18n^{2} + 42n - 36

This is not nice ... so, if there is an integer value for a solution, it must be a factor of 36 ...

keep trying them (1, -1, 2, -2, 3, -3, 4, -4, 6, -6, 9, -9, 18, -18, 36, -36) in the problem

until you find one that works ... fortunately, there is one: n = 6.

Factor out the (n - 6) factor to get (n - 6)(2n^{2 }- 6n + 6) = 2(n - 6)(n^{2} - 3n + 3)

n^{2} - 3n + 3 has no integer solutions (it doesn't even have real solutions),

so the only integer solution is** 6.**

geno3141 May 22, 2020