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

0
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Find all integers  for which
$$n^3 = (n-1)^3+(n-2)^3+(n-3)^3.$$

Sep 13, 2020

#1
+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
+339
+1

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

Nacirema  Sep 21, 2020
#2
+112053
+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
+339
+1

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