Proving that any rational number can be represented as the sum of the cubes of three rational numbers I found the following question in a book:
Prove that any integer can be represented as the sum of the each cube of five integers.
The answer : 
$$n=n^3+\left[-\frac{(n-1)n(n+1)}{6}-1\right]^3+\left[-\frac{(n-1)n(n+1)}{6}-1\right]^3+\left[\frac{(n-1)n(n+1)}{6}\right]^3+\left[\frac{(n-1)n(n+1)}{6}\right]^3.$$
This book says, "It is known that any rational number can be represented as the sum of the each cube of three rational numbers" without any additional information. I've tried to prove this, but I'm facing difficulty. Then, here is my question.
Question : Could you show me how to prove that any rational number can be represented as the sum of the each cube of three rational numbers.
I need your help.
 A: Theorem (Ryley 1825) Every rational number $R$ can be represented as the sum of three rational cubes:
$$
R=x^3+y^3+z^3.
$$
A short proof can be found in the artcle of Richmond (1930) here.
A: As a summary, what is known is that for any non-zero rational $N$, then
$$x_1^3+x_2^3+x_{\color{blue}3}^3 = N$$
$$y_1^5+y_2^5+\dots+y_{\color{blue}6}^5 = N$$
$$z_1^7+z_2^7+z_3^7+\dots+z_{\color{blue}8}^7 = N$$
where the variables are rationals. The case $p=3$ is by Ryley, while $p=5,7$ is by Choudhry. 
Since the link given in the other answer has expired, we can provide an explicit identity, 
$$(27m^3-n^9)^3 + (-27m^3+9mn^6+n^9)^3 + (27m^2n^3+9mn^6)^3 = m(27m^2n^2 +9mn^5+3n^8)^3$$
which is a simpler version by Robert Israel. The cases $p=5,7$ (and others) are discussed as Waring-like problems here, while there is also a detailed discussion of the elliptic curve involved for $p=7$ in this MSE post.
A: This fact follows immediately from the identity
$$
  ab^2 = \biggl(\frac{(a^2+3b^2)^3+(a^2-3b^2)(6ab)^2}{6a(a^2+3b^2)^2}\biggr)^{\!3}
+ \biggl(\frac{(3b^2-a^2)6ab^2}{(a^2+3b^2)^2}\biggr)^{\!3}
+ \biggl(\frac{(a^2+3b^2+6ab)(6ab-a^2-3b^2)}{6a(a^2+3b^2)}\biggr)^{\!3}.
$$
