Can any integer can be written as the sum of 8 integer cubes? Can anyone offer an elementary proof why: $$\forall n \in {\mathbb Z} \space \exists a, b, c, d, e, f, g, h \in {\mathbb Z}$$such that$$n=a^3+b^3+c^3+d^3+e^3+f^3+g^3+h^3{\rm ?}$$
In other words, why every integer is the sum of eight cubes integers.
My first thought was that since the gaps between $p^3$ and $(p+1)^3$ increase with $p,$ this would only hold for small integers.  This is wrong; can anyone tell me why?
N.B. This is different to Waring's problem because $n,a,b,c...$ dot not have to be natural numbers, so we can have $27=3^3+(-1)^3+(-1)^3+(-1)^3+(-1)^3$
 A: Allowing negative cubes, five suffice. It is suspected that four suffice but this is an open problem. I will see if I can find the argument for five, it is just one or two explicit formulas. By the way, these are called the "Easier" Waring problems. 
Can't find them at this point, so here is section D5 form Richard K. Guy, Unsolved Problems in Number Theory. Note that he is allowing cubes to be positive, negative, or zero: 

Is every number the sum of four cubes? This has been proved for all
  numbers except possibly those of the form $9n \pm 4$

Part 1: $$ 6n = (n+1)^3 + (n-1)^3 - n^3 - n^3  $$
Part 2:
$$ 6n - 2 = n^3 +(n+2)^3 - (n+1)^3 - (n+1)^3 - 2^3  $$ 
$$ 6n-1 =  (n+1)^3 + (n-1)^3 - n^3 - n^3 - 1^3  $$
$$ 6n = (n+1)^3 + (n-1)^3 - n^3 - n^3  $$
$$ 6n+1 =  (n+1)^3 + (n-1)^3 - n^3 - n^3 + 1^3  $$
$$ 6n + 2 = n^3 +(n-2)^3 - (n-1)^3 - (n-1)^3 + 2^3  $$
$$ 6n + 3 = (n-3)^3 +(n-5)^3 - (n-4)^3 - (n-4)^3 + 3^3  $$
A: *

*Since $n - n^3 = n(1-n)(1+n)$ then $6 \mid (n - n^3)$ and we can write $n - n^3 = 6k$.

*$6k = (k+1)^3 + (k-1)^3 - k^3 - k^3$.

*From 1 and 2 we derive that $n = n^3 + (k+1)^3 + (k-1)^3 - k^3 - k^3$ for $k=(n-n^3)/6$.
