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Find all integers  for which
\(n^3 = (n-1)^3+(n-2)^3+(n-3)^3\)

 May 22, 2020
 #1
avatar+744 
0

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

 May 22, 2020
 #2
avatar+21951 
0

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)(2n- 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.

 May 22, 2020

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