નિશ્ચાયકનું વિસ્તરણ કર્યા સિવાય સાબિત કરો : $\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right|$

$L H S .=\left|\begin{array}{lll}a & a^{2} & b c \\ b & b^{2} & c a \\ c & c^{2} & a b\end{array}\right|$

$=\frac{1}{a b c}\left|\begin{array}{lll}a^{2} & a^{3} & a b c \\ b^{2} & b^{3} & a b c \\ c^{2} & c^{3} & a b c\end{array}\right|$  $\left[R_{1} \rightarrow a R_{1}, R_{2} \rightarrow b R_{2}, \text { and } \mathrm{R}_{3} \rightarrow c R_{3}\right]$

$=\frac{1}{a b c} a b c\left|\begin{array}{lll}a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1\end{array}\right| \quad\left[\text { Taking out factor } a b c \text { from } C_{3}\right]$

$=\left|\begin{array}{lll}a^{2} & a^{3} & 1 \\ b^{2} & b^{3} & 1 \\ c^{2} & c^{3} & 1\end{array}\right|$

$=\left|\begin{array}{lll}1 & a^{2} & a^{3} \\ 1 & b^{2} & b^{3} \\ 1 & c^{2} & c^{3}\end{array}\right| \quad\left[\text { Applying } C_{1} \leftrightarrow C_{3} \,a n d\, C_{2} \leftrightarrow C_{3}\right]$

$=\mathrm{R} . \mathrm{H.S}$

Hence, the given result is proved.