If the two equations $a_1{x}^2+b_{1}x+c_1=0$  and $a_2{x}^2+b_{2}x+c_2=0$  have a common root, then the value of $\left(a_1{b}_2-a_2{b}_1\right)$ $\left(b_1{c}_2-b_2{c}_1\right)=$

The given equations are $a_1{x}^2+b_{1}x+c_1=0$

And $a_2{x}^2+b_{2}x+c_2=0$

Given that $\alpha$  is a common root of  Equations

Then the equations are $a_1{\alpha }^2+b_1\alpha +c_1=0$ ….. ( 1 )

${\alpha }_2{\alpha }^2+b_2\alpha +c_2=0$ ….. ( 2 )

Solve  ( 1 )&( 2 )

$\Rightarrow \frac{{\alpha }^2}{b_1{c}_2-c_1{b}_2}=\frac{\alpha }{c_1{a}_2-a_1{c}_2}=\frac{1}{a_1{b}_2-a_2{b}_1}$

$\Rightarrow \frac{{\alpha }^2}{\alpha }=\frac{b_1{c}_2-c_1{b}_2}{c_1{a}_2-a_1{c}_2}$   and $\frac{\alpha }{1}=\frac{c_1{a}_2-a_1{c}_2}{a_1{b}_2-a_2{b}_1}$

$\Rightarrow \frac{b_1{c}_2-c_1{b}_2}{c_1{a}_2-a_1{c}_2}=\frac{c_1{a}_2-a_1{c}_2}{a_1{b}_2-a_2{b}_1}.$

$\Rightarrow \left(a_1{b}_2-a_2{b}_1\right)$ $\left(b_1{c}_2-b_2{c}_1\right)=$ $\left(c_1{a}_2-a_1{c}_2\right)\left(c_1{a}_2-a_1{c}_2\right)$

$\therefore \left(a_1{b}_2-a_2{b}_1\right)$ $\left(b_1{c}_2-b_2{c}_1\right)=$ ${\left(a_1{c}_2-c_1{a}_2\right)}^2$