If the equation ${\mathrm{x}}^2-3\mathrm{px}+2\mathrm{q}=0\text{ and }{\mathrm{x}}^2-3\mathrm{ax}+2\mathrm{b}=0$  have a common roots and the other roots of the second equation
is the reciprocal of the other roots of the first, then $(2\mathrm{q}-2\mathrm{b}{)}^2$

According to the question

$\mathrm{\alpha }, \mathrm{\beta }$ are roots of equation ${\mathrm{x}}^2-3\mathrm{px}+2\mathrm{q}=0$

and $\mathrm{\alpha },\frac{1}{\mathrm{\beta }}$ are roots of equation ${\mathrm{x}}^2-3\mathrm{ax}+2\mathrm{b}=0$

Therefore, we have $\mathrm{\alpha }+\frac{1}{\mathrm{\beta }}=3\mathrm{a},\frac{\mathrm{\alpha }}{\mathrm{\beta }}=2\mathrm{b}$ 

and $\mathrm{\alpha }+\frac{1}{\mathrm{\beta }}=3\mathrm{a},\frac{\mathrm{\alpha }}{\mathrm{\beta }}=2\mathrm{b}$

$\begin{array}{r}\therefore (2\mathrm{q}-2\mathrm{b}{)}^2={\left(\mathrm{\alpha \beta }-\frac{\mathrm{\alpha }}{\mathrm{\beta }}\right)}^2={\mathrm{\alpha }}^2{\left(\mathrm{\beta }-\frac{1}{\mathrm{\beta }}\right)}^2\\ =\frac{\mathrm{\alpha }}{\mathrm{\beta }}\cdot \mathrm{\beta \alpha }{\left[(\mathrm{\alpha }+\mathrm{\beta })-\left(\mathrm{\alpha }+\frac{1}{\mathrm{\beta }}\right)\right]}^2\\ =\left(2\mathrm{b}\right)\left(2\mathrm{q}\right)[3\mathrm{p}-3\mathrm{a}{]}^2\\ =36\mathrm{bq}(\mathrm{p}-\mathrm{a}{)}^2\end{array}$