Help Pls, with explanation as well

n=6,3+i√32,3−i√32n=6,3+i32,3-i32, do you mean solve for n?

A formula for (a + b)³  =  a³ + 3·a²·b + a·b² + b³.

[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  =  a³   + 3·a²·b       + 3·a·b²       + b³.

(n - 1)³  =  (n)³ + 3·(n)²·(-1) + 3·(n)·(-1)² + (-1)³.

             =  n³ - 3n² + 3n - 1

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

(a + b)³  =  a³   + 3·a²·b       + 3·a·b²       + b³.

(n - 1)³  =  (n)³ + 3·(n)²·(-2) + 3·(n)·(-2)² + (-2)³.

             =  n³ - 6n² +12n - 8

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

(a + b)³  =  a³   + 3·a²·b       + 3·a·b²       + b³.

(n - 3)³  =  (n)³ + 3·(n)²·(-3) + 3·(n)·(-3)² + (-3)³.

             =  n³ - 9n² +27n - 27

Putting all this together:

n³  =  (n - 1)³ + (n - 2)³ + (n - 3)³

n³  =  ( n³ - 3n² + 3n - 1 ) + ( n³ - 6n² +12n - 8 ) + ( n³ - 9n² +27n - 27 )

n³  =  3n³ -18n² + 42n - 36

  0  =  2n³ - 18n² + 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)(2n² - 6n + 6) =  2(n - 6)(n² - 3n + 3)

n² - 3n + 3  has no integer solutions (it doesn't even have real solutions),

so the only integer solution is 6.