how to parameterize diophant triples on the elliptic curve? How to parameterize Diophantine triples on the elliptic curve?
Example: In the article of the Andrej Dujella , 
The parameterized state of the diophantine triples $(a, b, c)$ on the 
$$E : y^2 = x^3 + (3t^4 − 21t^2 + 3)x^2+(3t^8 +12t^6 +18t^4 +12t^2 +3)x+(t^2+1)^6$$
is
$$a = \dfrac{18t(t−1)(t+1)}{(t^2 − 6t + 1)(t^2 + 6t + 1)}\\ 
b=\dfrac{(t−1)(t^2 +6t+1)^2}{6t(t+1)(t^2 −6t+1)}\\ c=\cdots$$
How are these parameters?

The idea in the paper 
A. Dujella, M. Kazalicki, M. Mikic and M. Szikszai, There are infinitely many rational Diophantine sextuples, Int. Math. Res. Not. IMRN 2017 (2) (2017), 490-508.
is to construct rational Diophantine triples $a,b,c$ (i.e. $ab+1$, $ac+1$, $bc+1$) such that 
on the induced elliptic curve $y^2=(x+ab)(x+ac)(x+bc)$ the point with $x$-coordinate equal to $1$ has order $3$. 
The condition can be written in the form 
$$ \sigma_2=(\sigma_1^2 \sigma_3^2-12\sigma_3^2-6\sigma_1 \sigma_3-3)/(4+4\sigma_3^2), $$
where $\sigma_1=a+b+c$, $\sigma_2=ab+ac+bc$, $\sigma_3=abc$. 
Inserting this in the condition that $(ab+1)(ac+1)(bc+1)$ is a perfect square, 
we get $1+\sigma_3^2$ is a square, i.e. $\sigma_3=\frac{t^2-1}{2t}$. 
The polynomial $X^3-\sigma_1X^2+\sigma_2X-\sigma_3$ should have rational roots, 
so its discriminant has to be a perfect square.
From this condition we got the quartic in $\sigma_1$
$$ (t^6-2t^4+t^2)\sigma_1^4+(23t^3-23t^5+t^7-t)\sigma_1^3+(-45t^2-45t^6+126t^4)\sigma_1^2
+(162t^5-162t^3-54t^7+54t)\sigma_1-27-54t^4+108t^2+108t^6-27t^8=w^2, $$
which is equivalent to the elliptic curve written in the question. 
The curve in question has an obvious point $R$ with $x$-coordinate equal to $0$. 
We compute the point $2R$, which has $x$-coordinate $-3/4(t^2-6t+1)(t^2+6t+1)$. 
Transfering it back to the quartic, we get 
$$ \sigma_1 = \frac{t^8+130t^6-390t^4+130t^2+1}{3(t-1)(t+1)(t^2-6t+1)(t^2+6t+1)t}. $$
If we insert obtained values of $\sigma_1,\sigma_2,\sigma_3$ in $X^3-\sigma_1X^2+\sigma_2X-\sigma_3=0$, 
we get three rational solutions for $X$, which are exactly $a,b,c$ written in the question. 
