Independent chances of $3$ events ${a\over{a+x}}$, ${b\over{b+x}}$, ${c\over{c+x}}$ Here's a problem from my probability textbook:

Of three independent events the chance that the first only should happen is a; the chance of the second only is $b$; the chance of the third only is $c$. Show that the independent chances of the three events are respectively$${a\over{a+x}}, \quad {b\over{b+x}}, \quad {c\over{c+x}},$$where $x$ is a root of the equation$$(a+x)(b+x)(c+x) = x^2.$$

Here's what I did. We have$$a = p_1(1 - p_2)(1 - p_3), \quad b = (1-p_1)p_2(1-p_3), \quad c = (1-p_1)(1-p_2)p_3.$$Without loss of generality let's consider $a$. We have$$p_1 = {a\over{(1 - p_2)(1 - p_3)}}.$$If we assume the result we want to show, then this equals$$p_1 = {a\over{\left(1 - {b\over{b+x}}\right)\left(1 - {c\over{c+x}}\right)}} = {a\over{{{x^2}\over{(b+x)(c+x)}}}} = {a\over{a+x}}.$$However, we assumed in part what we wanted to show, which possibly makes this circular.
Another observation I noticed is that$$p_1p_2p_3 + p_1p_2(1-p_3) + p_1(1-p_2)p_3 + (1-p_1)p_2p_3 + (1-p_1)(1-p_2)(1-p_3) + a + b + c = 1.$$However, despite what I've tried, I'm stuck and do not know how to proceed further. Could anybody help me? Is there a way to turn my circular approach into a noncircular one?
 A: $$\begin{array}{rcl}a&=&p(1-q)(1-r)\\b&=&(1-p)q(1-r)\\c&=&(1-p)(1-q)r\end{array}$$
(where $p,q,r$ are the probabilities of those events. I prefer to use $p,q,r$ rather than $p_1, p_2, p_3$ - for the ease of typing!)
Now, let $x=(1-p)(1-q)(1-r)$ - the probability that none of the events happened. Then, obviously:
$$\begin{array}{rcl}\frac{a}{a+x}&=&\frac{p(1-q)(1-r)}{p(1-q)(1-r)+(1-p)(1-q)(1-r)}\\&=&\frac{p(1-q)(1-r)}{(1-q)(1-r)}\\&=&p\end{array}$$
and similarly for $q$ and $r$. Moreover:
$$\begin{array}{rcl}(a+x)(b+x)(c+x)&=&(1-q)(1-r)\cdot(1-p)(1-r)\cdot(1-p)(1-q)\\&=&(1-p)^2(1-q)^2(1-r)^2\\&=&x^2\end{array}$$
A: Note that you want $p_1=\frac{a}{a+x}$. Since $a=p_1(1-p_2)(1-p_3)$, then $a+x=(1-p_2)(1-p_3)$. You can further simplify $x=(1-p_1)(1-p_2)(1-p_3)$. This satisfies the equation in the question.
A: To get corner cases out of the way, if $p_1=1$ then $b=c=0$, and $x=1-a$ will satisfy all the wanted properties. If $p_1=0$, then $a=0$ and the problem reduces to a similar one with only two events. The same goes for $p_2$ and $p_3$, so from here on assume $0<p_1<1$, $0<p_2<1$, and $0<p_3<1$.
From your observation
$$a = p_1(1 - p_2)(1 - p_3), \quad b = (1-p_1)p_2(1-p_3), \quad c = (1-p_1)(1-p_2)p_3$$
we can get
$$ (1-p_1)(1-p_2)(1-p_3) = \frac{1-p_1}{p_1} a = \frac{1-p_2}{p_2} b = \frac{1-p_3}{p_3} c $$
If we call this value $x = (1-p_1)(1-p_2)(1-p_3)$, then $x>0$ and
$$ x = \frac{1-p_1}{p_1} a $$
$$ \frac{x}{a} = \frac{1}{p_1} - 1 $$
$$ \frac{1}{p_1} = \frac{a+x}{a} $$
$$ p_1 = \frac{a}{a+x} $$
Solving for $p_2$ and $p_3$ works just the same.
To show that $(a+x)(b+x)(c+x) = x^2$, it will be easier to look at the left side divided by $x^3$:
$$ \begin{align*} \frac{(a+x)(b+x)(c+x)}{x^3} &= \left(1+\frac{a}{x}\right) \left(1+\frac{b}{x}\right) \left(1+\frac{c}{x}\right) \\
 &= \left(1+\frac{p_1}{1-p_1}\right) \left(1+\frac{p_2}{1-p_2}\right) \left(1+\frac{p_3}{1-p_3}\right)  \\
&= \frac{1}{(1-p_1)(1-p_2)(1-p_3)} = \frac{1}{x}
\end{align*} $$
Now just multiply both sides by $x^3$ to get the polynomial equation in the question.
A: Let $\alpha$ be a root of $(a+x)(b+x)(c+x)=x^2$
$\implies (a+\alpha)(b+\alpha)(c+\alpha)=\alpha^2$
Now, $(p_1, p_2, p_3)$ satisfies the system of equations
$\begin{align}
(1) \quad p_1(1 - p_2)(1 - p_3) = a \\
(2) \quad (1-p_1)p_2(1-p_3) = b \\
(3) \quad (1-p_1)(1-p_2)p_3 = c
\end{align}$
It's sufficient to show that $\left(\frac{a}{a+\alpha},\frac{b}{b+\alpha}, \frac{c}{c+\alpha}\right)$ satisfies the above system of equations.
$(1) \quad \frac{a}{a+\alpha}\left(1 - \frac{b}{b+\alpha}\right)\left(1 - \frac{c}{c+\alpha}\right)=\frac{a}{a+\alpha}.\frac{\alpha}{b+\alpha}.\frac{\alpha}{c+\alpha}= \frac{a.\alpha^2}{(a+\alpha)(b+\alpha)(c+\alpha)} = \frac{a.\alpha^2}{\alpha^2}=a, \quad$ assuming $\alpha \neq 0$
$(2) \quad \left(1 - \frac{a}{a+\alpha}\right)\frac{b}{b+\alpha}\left(1 - \frac{c}{c+\alpha}\right)=\frac{\alpha}{a+\alpha}.\frac{b}{b+\alpha}.\frac{\alpha}{c+\alpha}= \frac{b.\alpha^2}{(a+\alpha)(b+\alpha)(c+\alpha)} = \frac{b.\alpha^2}{\alpha^2}=b, \quad$ assuming $\alpha \neq 0$
$(3) \quad \left(1 - \frac{a}{a+\alpha}\right)\left(1 - \frac{b}{b+\alpha}\right)\frac{c}{c+\alpha}=\frac{\alpha}{a+\alpha}.\frac{\alpha}{b+\alpha}.\frac{c}{c+\alpha}= \frac{c.\alpha^2}{(a+\alpha)(b+\alpha)(c+\alpha)} = \frac{c.\alpha^2}{\alpha^2}=c, \quad$ assuming $\alpha \neq 0$
