Has this equation appeared before? I want to know if the following equation has appeared in mathematical literature before, or if it has any important significance.
$$\sqrt{\frac{a+b+x}{c}}+\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{c+a+x}{b}}=\sqrt{\frac{a+b+c}{x}},$$
where $a,b,c$ are any three fixed positive real and $x$ is the unknown variable.
 A: 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.
A: I edit this answer to explicit the degree of the resulting polynomial having $x$ as a root (I wonder about the exact origin of this strange unknown x. Explanation is tedious and it will be presented abbreviated as possible, just to have the degree of the polynomial without using algebraic number theory which is not so useful without knowing the particular values of $a,b,c$).
First at all one has  $$\sqrt{\frac{b+c+x}{a}}+\sqrt{\frac{c+a+x}{b}}=\sqrt{\frac{a+b+c}{x}}-\sqrt{\frac{a+b+x}{c}}$$
Squared twice and, for convenience, make the following notations:
$$\begin {cases}A=\frac{a+b+x}{c}+\frac{a+c+x}{b}\\B=\frac{a+b+c}{x}+\frac{b+c+x}{a}\\M=(\frac{a+b+x}{c})(\frac{a+c+x}{b})\\N=(\frac{a+b+x}{x})(\frac{b+c+x}{a})\\C=\frac{B-A}{2}\\D=\frac{B^2-A^2+4N-4M}{4}\end{cases}$$ 
One has the linear system
$$\begin{cases}\sqrt M+\sqrt N=C\\ A\sqrt M+B\sqrt N=D\end{cases}\qquad(*)$$
So, for instance $$\sqrt M=\frac {CB-D}{B-A}\Rightarrow M=\left(\frac{CB-D}{B-A}\right)^2\qquad(**)$$
Thus a rational resultant is 
$$\left(\frac{a+b+x}{c}\right)\left(\frac{a+c+x}{b}\right)\left(B-A\right)^2=\left(B^2+A^2+4M-2AB-4N\right)^2\qquad (***) $$
Explaining $(***)$ with values of $a, b, c, x$ gives
$$G(x,a,b,c)-(H(x,a,b,c))^2=0$$ with 
$$G(x,a,b,c)=bc(ax)^2(x^2+A_1x+A_2)(A_3+x)^2(abc-A_4x)^2$$
$$H(x,a,b,c)=\{bc(x^2+(b+c)x+A_5)\}^2+\{ax((b+c)x+A_6)\}^2-2abcx\{(b+c)x+A_7)(x^2+(b+c)x+A_8)+2(ax^2+A_9x-A_{10})\}$$
where $$\begin{cases}A_1=2a+b+c\\ A_2=a^2+ab+ac+bc\\ A_3=a+b+c\\ A_4=ab+ac-bc\\ A_5=a(a+b+c)\\  A_6=b(a+b)+c(a+c)\\  A_7=A_6\\  A_8=A_5\\A_9=a^2+ac-bc\\ A_{10}=b^2c+bc^2\end{cases}$$
Therefore one can see at first sight that degree of $G$ is $8$ and so is for the degree of $H^2$ so that the degree of the resultant equation could be in general $8$.
►However it could be maybe for particular values of $a,b,c$ that the coefficients of $x^8$ in $G$ and $H^2$ be equal so the degree were less than $8$; this happen if 
$$bca^2(ab+ac-bc)^2=[b^2c^2+4a^2+a^2(b+c)^2(1-4b^2c^2)]^2$$ which seems to be possible for some values of $a,b,c$.
►On the other hand, solving also in $(*)$ for $N$, one gets $$N=\left(\frac{D-AC}{B-A}\right)^2$$ we can, in general, by multiplication and subtraction of the coefficients of $x^8$, eliminate the degree $8$ so this degree can be reduced.
►With the two polynomials giving by the system $(*)$, we can eliminate the unknown $x$ and find a relation $F(a,b,c)=0$ in order both polynomials have common roots (this can be made by Sylvester’ method, for instance, which would give a large zero determinant for $F(a,b,c)$
I stop here. I have wanted to help @pritam, assuming he is really interested in the rare equation proposed (I don’t discard some typo in my answer).
