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?

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 $\alpha \neq 0$
