Extremely Large Cubes

$5385624408 \equiv 0 (mod 9)$

$5385624408^3 \equiv 0 (mod 9)$

$3029229091 \equiv 1 (mod 9)$

$3029229091^3 \equiv 1 (mod 9)$

$7945317655 \equiv 7 (mod 9)$

$7945317655^3 \equiv 343 \equiv 1 (mod 9)$

$3151181443 \equiv 4 (mod 9)$

$3151181443^3 \equiv 64 \equiv 1 (mod 9)$

Plugging these values back to the question,

$(5385624408^3+3029229091^3)^3 + (7945317655^3+3151181443^3)^3$

$\equiv (0+1)^3 + (1+1)^3 (mod 9)$

$\equiv 1+8 (mod 9)$

$\equiv 9 (mod 9)$

$\equiv 0 (mod 9)$ Therefore, that number ( which I am too lazy to copy and paste it) is a multiple of 9. So, the digit sum of it must be a multiple of 9. Since the hint said that it was a one-digit number, the answer is 9. It can't be zero or else the number itself will be zero.

As $f(n)$ be the sum of the digits n we can easily derived some of the properties of $f(n)$ .Like $f(f(a+b))=f(f(a))+f(f(b))$ . To check this property you can put any two integer in place of a&b. Let , consider a=12 &b=15 then $f(f(a+b))=9$ and also $f(f(a))+f(f(b))=9$ So the property is correct. Another property which I use to solve this one is $f(f{(a^{3}+b^{3})^{3}})=f([f(a+b)]^{3})$ . These properties are found by observation. Then consider the given as $f[f((a^{3}+b^{3})^{3}+(c^{3}+d^{3})^{3})]$ . Then break it like $f[f(a^{3}+b^{3})^{3}]+f[f(c^{3}+d^{3})^{3}]$ and then use the second property which I consider before. You can get the answer $1+8=9$ .

If a number is divisible by 9, then the sum if its digits is divisible by 9; conversely, if the sum of the digits of a number is divisible by 9, then the number is divisible by 9 (this is a common di visibility test: notice that 1, 10, 100, etc. are all congruent to 1 mod 9, so the sum of the digits of a number base 10 is congruent to the number itself mod 9). Now we notice that each of the four numbers in the expression is divisible by 9, therefore the entire expression inside the inner f(...) is divisible by 9. We apply f twice to that number to get the final answer of 9 - we know the final answer can't be zero (since only zero has a digit sum of zero), so we use the hint to deduce that the answer is 9 and not a greater multiple of 9. (I have no proof that the final answer can't be 18 or greater, unfortunately :( )

luck by chance!!

I evaluated each expression by hand.I know this sounds stupid but it is the best I could do and I guess I got lucky :)