Nontrivial integer solutions of $ a^3+b^3=c^3+d^3$ How can we obtain a set of nontrivial solutions of

$$
a^3+b^3=c^3+d^3,
$$
  for $a,b,c,d\in \mathbb{Z}$
  where $(a,b)\neq (c,d)$ and $(a,b)\neq (d,c)$.



*

*Say in the range that $|a|,|b|,|c|,|d| \in [0,30]$, whose absolute values small or eqaul to 30?

*How many simple but nontrivial solutions are there around this range? 
The simple form the solutions are the better. 
Thank you!
p.s. When $d=0$, we know it is impossible due to the Fermat's Last Theorem.
 A: It is possible to give a partial answer by mimicking Kummer's approach to the first case of FLT. As in your other related question, let us introduce $\omega$ = a primitive 3rd root of unity, the quadratic field $\mathbf Q(\omega)$, its ring of integers $\mathbf Z[\omega]$. We intend to solve the diophantine equation $a^3+b^3=c^3+d^3$ under the additional hypothesis that $3\nmid a+b$ (or equivalently $3\nmid c+d$, by Fermat's little theorem). For simplification, let us concentrate on primitive solutions, i.e. suppose that $a, b$ are coprime (and similarly $c,d$ ).To exploit the decomposition $a^3+b^3=(a+b)(a+b\omega)(a+b\omega^2)$, we must recall that $\mathbf Z[\omega]$ is a principal ideal ring, whose group of units coincides with its group of roots of unity, generated by $\pm \omega$. 
Lemma: Under our hypotheses, for $m\neq  n$ mod 3, the factors $(a+b\omega^m)$ and $(a+b\omega^n)$ are coprime in $\mathbf Z[\omega]$. Proof: Since $\mathbf Z[\omega]$ is principal, let us apply Bézout's theorem by showing that  the ideal $J$ generated by our two elements contains $1$. Because $\omega^k$ is a primitive 3rd root of unity whenever $k \neq 0$ mod 3, the quotient $\epsilon_k=\frac {1-\omega^k}{1-\omega}$ is a unit. But $(a+b\omega^m)-(a+b\omega^n)=\omega^m(1-\omega^{n-m})b=\epsilon_{n-m}\omega^m(1-\omega)b$ , and similarly $(a+b\omega^m)\omega^n-(a+b\omega^n)\omega^m=-\omega^m(1-\omega^{n-m})a= - \epsilon_{n-m}\omega^m(1-\omega)a$, hence $(1-\omega)b$ and $ (1-\omega)a \in J$. Since $a,b$ are coprime in $\mathbf Z$,  Bézout asserts the existence of $u,v \in \mathbf Z$ s.t. $ua+vb=1$, hence $(1-\omega)ua+(1-\omega)vb=(1-\omega) \in J$, hence also $3\in J$. Moreover, $a+b=(a+b\omega^m)+(1-\omega^m)b=(a+b\omega^ma+b\omega^m)+(1-\omega)\epsilon_mb$, so that $a+b \in J$. Finally, our additional hypothesis implies the existence of $s,t\in \mathbf Z$ s.t. 
$(a+b)s+3t=1$, hence $1\in J$. OUF
EDIT : Any rational prime $p\neq 3$ is unramified in $\mathbf Q(\omega)$. Let $\delta$= gcd $(a+b,c+d)$ and $N$= the norm of $\mathbf Q (\omega)/\mathbf Q$ . Our additional hypothesis and the lemma then imply that  $\frac {a+b}{\delta}=\pm N(c+d\omega)=\pm (c^2+d^2-cd)=\pm ((c+d)^2-3cd)$, and similarly $\frac {c+d}{\delta}=\pm ((a+b)^2-3ab)$. Besides, since $c,d$ are the roots of the quadratic equation $x^2-(c+d)x+cd=0$, the discriminant $\Delta=(c+d)^2-4cd=((c+d)^2-3cd)-cd$ is necessarily a perfect square (=square of an integer), and similarly $((a+b)^2-3ab)-cd$ is a perfect square. Summarizing, $\pm\frac{a+b}{\delta}-cd$ and $\pm\frac{c+d}{\delta}-ab$ must be perfect squares. 
EDIT 2: I made a mistake again ! $(a+b\omega)$ and $(c+d\omega)$ could have a common factor. This seems to be a dead end.
