Has this equation appeared before?

This provides the explicit polynomial in $x$ (for those curious), though I'm not aware if the equation has appeared in the mathematical literature. We get rid of the square roots by multiplying out the $8$ sign changes,

$$\prod^8 \left(\sqrt{\frac{a+b+x}{c}}\pm\sqrt{\frac{b+c+x}{a}}\pm\sqrt{\frac{a+c+x}{b}}\pm\sqrt{\frac{a+b+c}{x}}\right)=0$$

then collecting powers of $x$. It turns out the $8$th-deg equation factors into a linear (cubed), a quadratic, and a cubic. For simplicity, let,

$$\begin{aligned} p &= a+b+c\\ q &= ab+ac+bc\\ r &= abc \end{aligned}$$

Then,

$$(p+x)^3=0\tag1$$

$$r^2 - 2 q r x + (q^2 - 4 p r) x^2 = 0\tag2$$

$$p r^2 + r (-2 p q + 9 r) x + (p q^2 - 4 p^2 r + 6 q r) x^2 + (q^2 - 4 p r) x^3 = 0\tag3$$

Example:

Let $a,b,c = 1,2,4$, then

$$(7+x)^3=0\\ -16 + 56 x + 7 x^2 = 0\\ -112 + 248 x - 119 x^2 + 7 x^3 = 0$$

The roots of the quadratic solve,

$$\sqrt{\frac{a+b+x}{c}}\pm\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{a+c+x}{b}}-\sqrt{\frac{a+b+c}{x}}=0$$

while a root of the cubic solves,

$$\sqrt{\frac{a+b+x}{c}}-\sqrt{\frac{b+c+x}{a}}-\sqrt{\frac{a+c+x}{b}}+\sqrt{\frac{a+b+c}{x}}=0$$

and two others, while the linear root takes care of the remaining three sign changes.