$(1-a^2)(1-b^2)(1-c^2)=8abc\; (a,b,c\in \mathbb{Q}^{+})$ 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?

 A: There are nontrivial rational solutions, for example $$(a,b,c)=(5/6,\ 2/11,\ 3/13).$$ Note that the equation may be written as 
$$[1]\ \ \frac{1-a^2}{2a}\cdot \frac{1-b^2}{2b} \cdot \frac{1-c^2}{2c}=1.$$
The following is only an idea of how to find an infinite number of pairs $(a,b)$ for which $(a,b,5/6)$ is a solution. Define $$f(x)=\frac{1-x^2}{2x},$$ and note that $f(p/q)=(q^2-p^2)/(2pq).$ Also note that if $f(p/q)=m/n$ then also $f((q-p)/(q+p))=n/m,$ so that we may invert one of the fractions in working with the equation.
If we can find infinitely many rational pairs $(x,y)$ for which
$$[2]\ \ \frac{1-x^2}{2x}\cdot \frac{2y}{1-y^2}=\frac{6}{5},$$ this will produce infinitely many rational solutions to [1], by the above remark about inverting fractions taken on by $f$  via the map $p/q \to (q-p)/(q+p).$
Now [2] is a cubic curve
$$kx-y+x^2y-ky^2=0,$$ where $k=6/5.$ It has a rational point on it taken from the solution $(5/6,2/11,3/13)$ to [1]. 
I think this means there are infinitely many rational points on [2]. From the single rational point $P$, use of the tangent line will give another, say $Q$; then use of the tangent line there will give a third point $R$, and after that one has two rational points $Q,R$ for which the curve is not tangent to line $QR$, so the line through those gives yet another point, etc. I am not expert enough on cubic curves to be able to definitely say this will give infinitely many rational points. Maybe someone who knows cubic curves can make this argument go through.
A: 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.
