Does there exist a function $f_{\Box,\Box}(\Box)$ making the formula $a + (b \oplus c) = (f_{b,c}(a)+b) \oplus (f_{c,b}(a)+c)$ true? Let $a$ and $b$ denote the resistances of two resistors. If they're put in series, the total resistance is $a+b$. If they're put in parallel, the total resistance is $$a \oplus b := \frac{1}{\frac{1}{a}+\frac{1}{b}} = \frac{ab}{a+b}.$$
I suspect there's a formula describing how $+$ "distributes over" $\oplus$.
It should be of the form
$$a + (b \oplus c) = (f_{b,c}(a)+b) \oplus (f_{c,b}(a)+c)$$
for an appropriate choice of $f$. I haven't been able to find an $f$ that works, however.

Question. Does there exist an $f$ making the above formula true? If not, why not?

Remark. Unpacking the definitions, we're looking for a function $f$ such that
$$(ab+ac+bc)(f_{b,c}(a)+b + f_{c,b}(a)+c)$$
and 
$$(b+c)(f_{b,c}(a)+b) (f_{c,b}(a)+c)$$
are equal.
 A: Let me assume that $b\ge0,c\ge0, b+c>0$.
I am looking for $f$ satisfying $f_{b,c}(a) = f_{c,b}(a)$.
The equation in the OP is then equivalent to
$$
(b+c) f_{bc}(a)^2 + ((b+c)^2-2ab-2bc-2ca)f_{bc}(a)- (b+c)(ab+bc+ca)=0.
$$
Setting $d:=ab+bc+ca$ and dividing by $b+c$ this is equivalent to 
$$
f_{bc}(a)^2 + (b+c -\frac{2d}{b+c}) f_{bc}(a)- d=0.
$$
The roots of this equation are 
$$\begin{split}
f_{bc}(a) &= -\frac{ (b+c)^2 - 2d}{2(b+c)} \pm \frac1{b+c} \sqrt{ \frac14((b+c)^2 - 2d)^2+d(b+c)^2}\\
&=-\frac{ (b+c)^2 - 2d \pm\sqrt{ (b+c)^4 +d^2}}{2(b+c)},
\end{split}
$$
one of them is positive, which gives you a formula for $f$.
A: There are many different possible definitions of $f_{b,c}(a)$. One approach is based on the identity $$(kb)\oplus (kc) = k(b\oplus c)$$
Since we would like to get $$a + \left(b\oplus c\right)=\left(f_{b,c}(a) + b\right)\oplus\left(f_{c,b}(a) + c\right)$$
it would be nice if the right-hand side can transformed into to contain $\left(b\oplus c\right)$. It can be rewritten as $$\left(f_{b,c}(a) + b\right)\oplus\left(f_{c,b}(a) + c\right) = 
\left(\left(\frac{f_{b,c}(a)}{b}+1\right)\cdot b\right)\oplus
\left(\left(\frac{f_{c,b}(a)}{c}+1\right)\cdot c\right)$$
Equating $\frac{f_{b,c}(a)}{b} = \frac{f_{c,b}(a)}{c}$ and setting both equal to $g(a,b,c)$ yields
$$a + \left(b\oplus c\right) = \left(g(a,b,c)+1\right)\left(b\oplus c\right)$$ Thus, we can set $g(a,b,c)=a/\left(b\oplus c\right)$, we get $$f_{b,c}(a)=\frac{ab}{b\oplus c}$$
and the equality 
$$a + \left(b\oplus c\right)=\left(\frac{ab}{b\oplus c} + b\right)\oplus\left(\frac{ac}{b\oplus c} + c\right)$$
Alternatively, we can also expand $\left(b\oplus c\right)=\frac{bc}{b+c}$ and get $$f_{b,c}(a)=\left(1+\frac{b}{c}\right)a$$
A: Defining
$$
f_{b,c}(a) = 2a+b-\frac{2b^2}{b+c}\ \ \text{or}\ \ \ f_{c,b}(a) = 2a+c-\frac{2c^2}{b+c}
$$
with $f_{b,c}(a)\ne f_{c,b}(a)$, we have
$$
a + (b \oplus c) = (f_{b,c}(a)+b) \oplus (f_{c,b}(a)+c)
$$
