+0  
 
0
98
5
avatar+14 

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

 Sep 13, 2020
 #1
avatar
+1

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

 

I could only find one value for n:

 

n = 6

 

6^3 = 5^3 + 4^3 + 3^3 = 216

 Sep 13, 2020
 #3
avatar+246 
0

Yes, but there are 2 more that you can't find by bashing it.

Nacirema  Sep 21, 2020
 #2
avatar+111127 
+1

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

 

\(n^3 = (n-1)^3+(n-2)^3+(n-3)^3\\ n^3 = (n^3-3n^2+3n-1)+(n^3-3*2n^2+3*4n-8)+(n^3-3*3n^2+3*9n-27)\\ n^3 = n^3-3n^2+3n-1+n^3-6n^2+12n-8+n^3-9n^2+27n-27\\ 0= 2n^3 -18n^2 +42n -36 \\ 0= n^3 -9n^2 +21n -18 \\\)

You can take if from there.

 Sep 20, 2020
 #4
avatar+246 
0

I got three solutions:

 

1. \(6\)

2. \(\frac {3+i\sqrt3}2\)

3. \(\dfrac {3-1\sqrt3}2\)

 

You can also write it as 

 

\(6, \dfrac {3\pm i\sqrt3}2\)

 Sep 21, 2020

55 Online Users