$(1-a^2)(1-b^2)(1-c^2)=8abc\; (a,b,c\in \mathbb{Q}^{+})$ has infinitely many solutions Prove that:
$$(1-a^2)(1-b^2)(1-c^2)=8abc,\quad a,b,c\in \mathbb{Q}^{+}$$
has an infinite number of solutions $(a,b,c)$.
Now we found this infinite 
$$(a,b,c)=(\dfrac{4p}{p^2+1},\dfrac{p^2-3}{3p^2-1},\dfrac{(p+1)(p^2-4p+1)}{(p-1)(p^2+4p+1)}),p>2+\sqrt{3},p\in\mathbb {Q}^{+}$$

My Question: Have other form solution?


The equation,
$$(1-a^2)(1-b^2)(1-c^2)-8abc=0$$
is just a quadratic in any of the variables. Hence,
$$c=\frac{8ab\pm y}{2(-1+a^2+b^2-a^2b^2)}$$
where,
$$64a^2b^2-4(-1+a^2+b^2-a^2b^2)(1-a^2-b^2+a^2b^2) = y^2$$
This quartic polynomial to be made a square is easily reducible to an elliptic curve. From initial rational point {$a,b$} = {$5/6,\; 2/11$}, one can find an infinite more.
