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If ${\Delta _1} = \left| {\begin{array}{*{20}{c}}
{{b^5}{c^6}\left( {{c^3} - {b^3}} \right)}&{{a^4}{c^6}\left( {{a^3} - {c^3}} \right)}&{{a^4}{b^5}\left( {{b^3} - {a^3}} \right)} \\
{{b^2}{c^3}\left( {{b^6} - {c^6}} \right)}&{a{c^3}\left( {{c^6} - {a^6}} \right)}&{a{b^2}\left( {{a^6} - {b^6}} \right)} \\
{{b^2}{c^3}\left( {{c^3} - {b^3}} \right)}&{a{c^3}\left( {{a^3} - {c^3}} \right)}&{a{b^2}\left( {{b^3} - {a^3}} \right)}
\end{array}} \right|$ and ${\Delta _2} = \left| {\begin{array}{*{20}{c}}
a&{{b^2}}&{{c^3}} \\
{{a^4}}&{{b^5}}&{{c^6}} \\
{{a^7}}&{{b^8}}&{{c^9}}
\end{array}} \right|$ then ${\Delta _1}{\Delta _2}$ is equal to
$\Delta _2^3$
$\Delta _2^2$
$\Delta _2^4$
None of these
Solution
$\Delta_{1}=\left(\Delta_{2}\right)^{n-1}\{n=3\}$
so, $\Delta_{1}=\Delta_{2}^{2}$
and $\Delta_{1} \Delta_{2}=\Delta_{2}^{2} \cdot \Delta=\Delta_{2}^{2}$
