Let's consider perfect cubes modulo 7:
a³ = b (mod 7)
0³=0, 1³=1, 2³=1, 3³=-1, 4³=1, 5³=-1, 6³=-1 and 7³=0 (all mod 7)
So perfect cubes leave remainders 0, 1 or -1 when divided by 7. In other words, perfect cubes are of the form:
7k+1, 7k-1 or 7k.
Now consider the sum of 3 cubes of the form 7k-1, it'd be of the form 7p-3=7n+4. So numbers of the form 7n+4 can be written as the sum of 3 perfect cubes. Let's check this for n={1,2,3}:
7*1+4 = 11 = 3³ + (-2)³ + (-2)³
7*2+4 = 18 = 3³ + (-2)³ + (-1)³
7*3+4 = 25 = 3³ + (-1)³ + (-1)³
Finally, let's consider 801,345,230,914 modulo 7:
801,345,230,914 = 4 (mod 7), which means it's of the form 7k+4 and can be written as the sum of 3 cubes.