Solve $\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} = -7$ for $x$ I have been scratching my head for solving this equation but I am unable to do this. Even I am unable to get how to use the hint. A way to solve this would be of great help

Solve for $x$ :-
$$\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} = -7$$
(Hint: $-3-4=-7$, if $\frac 1a+\frac 1b +\frac 1c \ne 0$)

 A: \begin{align}
\frac{a+b-x}c + \frac{a+c-x}b + \frac{b+c-x}a -\frac{4abc}{a+b+c} &= -7\\
\frac{a+b-x}c+\frac{c}{c} + \frac{a+c-x}b+\frac{b}{b} + \frac{b+c-x}a+\frac{a}{a} -\frac{4abc}{a+b+c} &= -7+3\\
\frac{a+b+c-x}c + \frac{a+b+c-x}b + \frac{a+b+c-x}a -\frac{4abc}{a+b+c} &= -4\\
(a+b+c-x)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)&=\frac{4abc}{a+b+c}-4\\
a+b+c-x&=\frac{4abc(abc-a-b-c)}{(a+b+c)(ab+bc+ca)}
\end{align}
So, $\displaystyle x=a+b+c-\frac{4abc(abc-a-b-c)}{(a+b+c)(ab+bc+ca)}$.
A: Hint: the equation looks scary with multiple $x$'s on the left hand side and all these division bars. However, from the perspective of finding $x$ it really is just of the form
$$A + Bx = -7$$
where $A$ and $B$ are some hideously complicated expressions in $a, b$ and $c$ (but not $x$!)
If you can solve $A + Bx = -7$, yielding a solution 

$x = $ some expression in $A$ and $B$

then you can get back to a solution in the original form with $a, b$ and $c$ by substituting in the right hand side the expressions for $A$ and $B$ you found. 
A: Let $a^{-1}+b^{-1}+c^{-1}=k$. Your equation becomes
$$
\frac{a}{c}+\frac{b}{c}+
\frac{a}{b}+\frac{c}{b}+
\frac{b}{a}+\frac{c}{a}-
\frac{4abc}{a+b+c}+7
=kx
$$
The first six terms on the left-hand side can be rewritten as
$$
\frac{a^2b+ab^2+a^2c+ac^2+b^2c+cb^2}{abc}
$$
Now, set $S=a+b+c$, $Q=ab+bc+ca$ and $P=abc$. The numerator of the above fraction can be computed as
$$
a^2b+ab^2+a^2c+ac^2+b^2c+cb^2=SQ-3P
$$
using the theory of symmetric polynomials. 
Basically, we must have
$$
a^2b+ab^2+a^2c+ac^2+b^2c+cb^2=\alpha S^3+\beta SQ+\gamma P
$$
for some constants $\alpha$, $\beta$ and $\gamma$; then
\begin{align}
a=1,b=0,c=0 &\implies 0=\alpha \\
a=1,b=1,c=0 &\implies 2=2\beta \\
a=1,b=1,c=1 &\implies 6=27\alpha+9\beta+\gamma
\end{align}
so $\alpha=0$, $\beta=1$ and $\gamma=-3$.
Since, easily, $k=Q/P$, the equation becomes
$$
\frac{Q}{P}x=\frac{SQ-3P}{P}-\frac{4P}{S}+7=
\frac{S^2Q-3PS-4P^2+7PS}{SP}=\frac{S^2Q+4PS-4P^2}{SP}
$$
and, finally,
$$
x=\frac{S^2Q+4PS-4P^2}{SQ}=S-\frac{4P(P-S)}{SQ}
$$
which agrees with the other solution.
