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Show that there are infinitely many rational triples $(a,b,c)$ such that $$a + b + c = abc = 6.$$

I first thought about proving that if a triangle $ABC$ has rational side lengths and rational area, then $\tan{\frac{A}{2}},\tan{\frac{B}{2}},\tan{\frac{C}{2}}$ are all rational. Thus, using the identity $\tan{2x} = \dfrac{2\tan{x}}{1-\tan^2{x}}$, we see that $\tan{A},\tan{B},\tan{C}$ are all rational. Thus since $$\tan{A}+\tan{B}+\tan{C} = \tan{A}\tan{B}\tan{C},$$ we just need $\tan{A}\tan{B}\tan{C} = 6$. How can we satisfy this or does this approach not work?

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    We want to find infinitely many $(x,y)\in\mathbb{Q}^2$ such that $(2+x)(2+y)(2-x-y)=6$ . The latter equation is that of a cubic ... Are there infinitely many rational points on it ? If a specialist of these questions has got an idea, it will be wellcome ... – 2017-01-20
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    @Adren If we fix $x$, then it's a quadratic in $y$. That means we need to show that there are infinitely many rational $x$ that makes the discriminant a (rational) square. – 2017-01-20
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    There are infinitely many points on the corresponding cubic. It is an elliptic curve of positive Mordel-Weil rank. – 2017-01-20
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    Related: http://math.stackexchange.com/questions/1384653/rational-solutions-to-abc-abc-6 – 2017-01-20
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    @Arthur : The question becomes finding if $x^4-4x^3-12x^2+8x+16$ is a rational square for infinitely many rational numbers $x$ ... which doesn't look simple to me. – 2017-01-20
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    @Adren I agree, but I think it's the natural way to continue on from your comment. Note that that polynomial at least has $x+2$ as a factor. – 2017-01-20
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    @Arthur Do you know if my approach can work? – 2017-01-20
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    After having a look at the link provided by @Byron Schmuland, I understand why I couldn't find it an easy question :) – 2017-01-20
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    So far, I only found $(1/2/3)$ , $(-3/2,-1/2,8)$ and $(25/21,49/15,54/35)$. Are there other solutions ? – 2017-01-20

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