If $\mathrm{f}\left(\mathrm{x}\right)=\mathrm{a}+\mathrm{bx}+{\mathrm{cx}}^2$ and $\mathrm{\alpha }, \mathrm{\beta }, \mathrm{\gamma }$ are the roots of the equation ${\mathrm{x}}^3=1\text{, then }\left|\begin{array}{l}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{c}\\ \mathrm{b}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{a}\\ \mathrm{c}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{a}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{b}\end{array}\right|$ is equal to

$\begin{array}{c}\left|\begin{array}{lll}\mathrm{a}& \mathrm{b}& \mathrm{c}\\ \mathrm{b}& \mathrm{c}& \mathrm{a}\\ \mathrm{c}& \mathrm{a}& \mathrm{b}\end{array}\right|=-\left({\mathrm{a}}^3+{\mathrm{b}}^3+{\mathrm{c}}^3-3\mathrm{abc}\right)\\ =-(\mathrm{a}+\mathrm{b}+\mathrm{c})\left(\mathrm{a}+{\mathrm{b\omega }}^2+\mathrm{c\omega }\right)\left(\mathrm{a}+\mathrm{b\omega }+{\mathrm{c\omega }}^2\right)\end{array}$

                               (where $\omega$ is cube roots of unity)

                   $=-\mathrm{f}\left(\mathrm{\alpha }\right)\mathrm{f}\left(\mathrm{\beta }\right)\mathrm{f}\left(\mathrm{\gamma }\right)\left[\because \mathrm{\alpha }=1,\mathrm{\beta }=\mathrm{\omega },\mathrm{\gamma }={\mathrm{\omega }}^2\right]$