If $\alpha ,\beta ;\beta ,\gamma ;\gamma ,\alpha$ are the roots of $a_i{x}^2+bx+c_i=0,i=1,2,3$ find the value of $(1+\alpha )(1+\beta )(1+\gamma )$

We have, 

$\begin{array}{r}\alpha +\beta =-\frac{b_1}{{\alpha }_1},\alpha \beta =\frac{c_1}{{\alpha }_1}\\ \beta +\gamma =-\frac{b_2}{a_2},\beta \gamma =\frac{c_2}{a_2}\end{array}$

and $\gamma +\alpha =-\frac{b_3}{a_3},\gamma \alpha =\frac{c_3}{{\alpha }_3}$ Now, $(1+\alpha )(1+\beta )=1+(\alpha +\beta )+\alpha \beta$

$\begin{array}{r}=1-\frac{b_1}{a_1}+\frac{c_1}{a_1}\\ =\frac{a_1-b_1+c_1}{a_1}\end{array}$.                 .. (i)

Similarly, $(1+\beta )(1+\gamma )=\frac{a_2-b_2+c_2}{a_2}$   ... (ii)

$(1+\gamma )(1+\alpha )=\frac{a_3-b_3+c_3}{a_3}$   …(iii)

Multiplying Eqs (i), (ii) and (iii), we get

$\begin{array}{l}\left\{\right(1+\alpha \left)\right(1+\beta \left)\right(1+\gamma ){\}}^2\\ =\left(\frac{a_1-b_1+c_1}{a_1}\right)\left(\frac{a_2-b_2+c_2}{a_2}\right)\left(\frac{a_3-b_3+c_3}{a_3}\right)\\ =\prod _{i=1}^3 \left(\frac{a_i-b_i+c_i}{a_i}\right)\end{array}$

Thus, $(1+\alpha )(1+\beta )(1+\gamma )={\left(\prod _{i=1}^3 \left(\frac{a_i-b_i+c_i}{a_i}\right)\right)}^{1/2}$