Proof of $a^3 - b^3 = c^3 + d^3$, where $a,b,c,d$ all rational?

Question

Reading Wikipedia article on Diophantus, it says in a book that survived that he makes reference to a lost book called Porisms and the theorem stated in the title: the difference between the cubes of any 2 rationals can be expressed as the sum of the cubes of 2 rationals. Anyone point me to this proof?

Answer 1

$c=\frac{a(a^3-2b^3)}{a^3+b^3},d=\frac{b(2a^3-b^3)}{a^3+b^3}$ gives a solution which I believe is due to Vieta.

Answer 2

The question asks about a theorem of Diophantus.

L. E. Dickson, History of the Theory of Numbers, Volume II, Chapter XXI, section "Two equal sums of two cubes", pp. $550$-$561$ begins on page $550$ with

$\quad$ Diophantus, V, $19$, mentions without details the theorem in the Porisms that the difference of two cubes is always a sum of two cubes (cf. p. $607$).

$\quad$ F. Vieta$^{38}$ required two cubes whose sum equals the difference $\,B^3-D^3\,$ of two given cubes $\,(B>D).\,$ Call $\,B-A\,$ the side of the first required cube and $\,B^2A/D-D\,$ the side of the second. Thus $\,(B^3+D^3)A=3D^3B\,$ and hence $$(1)\qquad x^3+y^3=B^3-D^3,\quad x=\frac{B(B^3-2D^3)}{B^3+D^3}, \quad y=\frac{D(2B^3-D^3)}{B^3+D^3}. $$

Following Vieta, let $\,b>a>0\,$ be two positive integers. Define

$$ x := b(b^3-2a^3),\;\; y := a(2b^3-a^3),\\ c := x/(b^3+a^3),\;\; d := y/(b^3+a^3). \tag{1} $$

The algebraic equality $$c^3 + d^3 = b^3 - a^3 \tag{2} $$

can be verified using elementary algebra by substituting the definitions from equation $(1)$ and expanding $\,c^3+d^3.\,$ Here$\,c>0\,$ requires that $\,b^3>2a^3\,$ which implies also $\,d>0.\,$

The two given numbers $\,a,b\,$ were arbitrary and equation $(2)$ holds which proves that the difference of two cubes is also the sum of two cubes. This answers the question

Anyone point me to this proof?

by giving a proof as requested.