Prove that the sum of digits of $(999...9)^{3}$ (cube of integer with $n$ digits $9$) is $18n$ Someone had posted a question on this site as to what would be sum of digits of $999999999999^3$ (twelve $9s$ ) equal to? 
I did some computation and found the pattern that sum of digits of  $9^3 = 18$,  $99^3 = 36$,  $999^3 = 54$ and so on. So, I had replied that sum of digits of  $999999999999^3 = 12 \cdot 18 = 216$.
Can anybody help me prove this, that sum of digits of  $(\underbrace{999\dots9}_{n\text{ times}})^3 = 18n.$
 A: With $n\geq 1$: 
$(10^n-1)^3=10^{3n}-3\cdot 10^{2n}+3\cdot 10^n -1$ $=(10^{n-1}-1)10^{2n+1}+7\cdot 10^{2n}+2\cdot 10^n+10^n-1$. 
Therefore $9(n-1)+7+2+9n=18n$.  
A: Observe that:
$\left(\underbrace{9\dots9}_{n\text{ times}}\right)^3=(10^n-1)^3=10^{3n}-3\cdot10^{2n}+3\cdot10^{n}-1=\underbrace{9\dots9}_{n-1\text{ times}}7\underbrace{0\dots0}_{n-1\text{ times}}2\underbrace{9\dots9}_{n\text{ times}}$

Therefore, the sum of digits is $9(n-1)+7+2+9n=18n$.
A: Well, all you have to do is to observe the pattern of the cubes themselves:
$9^3 = 729$.
$99^3=970299$
$999^3=997002999$
$9999^3=999700029999$
Now, you are in a position to make a conjecture:

Let $(k)_n$ mean the number in base $10$ represented by $k$ repeated $n$ times. Then, $((9)_n)^3 = (9)_{n-1}7(0)_{n-1}2(9)_n$.

I want you to go out and prove this conjecture yourself, use the fact that $(9)_n = 10^{n}-1$, and $(10^{n} -1 )^3 = 10^{3n} - 3\cdot 10^{2n} + 3 \cdot 10^n - 1$.
Now, use the fact that the sum of digits of $(k)_n$ is $nk$. Putting this formula above, the sum of digits of $((9)_n)^3$ is $9(n-1) + 7 + 2 + 9(n) = 18n$. Hence, for $12$ digits, your formula gives a sum of $216$, which is correct. 
A: The cube of '9 is '72'9, where 'x is a digit before the base just before 10, so '7 = 10 -3 etc.
The sum of digits in the general base is 1'8 (ie 2,0 - 2)
When taken for the general case of 9(n)9³ = 9(n)7 0(n)2 9(n)9, it is 18n+18, but because we mean n to be the number of digits in the lead, 18(n).
