નિશ્ચાયકના ગુણધર્મોનો ઉપયોગ કરીને સાબિત કરો કે:
$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$
$\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3} \end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)$
(A) ધારો કે $\Delta=\left|\begin{array}{lll}1 & 1 & 1 \\ a & b & c \\ a^{3} & b^{3} & c^{3}\end{array}\right|$.
$C_{1} \rightarrow C_{1}-C_{2}$ અને $C_{2} \rightarrow C_{2}-C_{3}$ લેતા:
$\Delta=\left|\begin{array}{ccc}0 & 0 & 1 \\ a-b & b-c & c \\ a^{3}-b^{3} & b^{3}-c^{3} & c^{3}\end{array}\right|$.
$C_{1}$ માંથી $(a-b)$ અને $C_{2}$ માંથી $(b-c)$ સામાન્ય લેતા:
$\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 1 & 1 & c \\ a^{2}+ab+b^{2} & b^{2}+bc+c^{2} & c^{3}\end{array}\right|$.
$C_{1} \rightarrow C_{1}-C_{2}$ લેતા:
$\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & c \\ a^{2}-c^{2}+ab-bc & b^{2}+bc+c^{2} & c^{3}\end{array}\right|$.
$\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & c \\ (a-c)(a+c)+b(a-c) & b^{2}+bc+c^{2} & c^{3}\end{array}\right|$.
$\Delta=(a-b)(b-c)(a-c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & c \\ a+b+c & b^{2}+bc+c^{2} & c^{3}\end{array}\right|$.
$R_{1}$ ની સાપેક્ષ વિસ્તરણ કરતા:
$\Delta=(a-b)(b-c)(a-c)(1)[0 - (a+b+c)] = -(a-b)(b-c)(a-c)(a+b+c)$.
કારણ કે $-(a-c) = (c-a)$,તેથી:
$\Delta=(a-b)(b-c)(c-a)(a+b+c)$.
આમ,પરિણામ સાબિત થાય છે.
$C_{1} \rightarrow C_{1}-C_{2}$ અને $C_{2} \rightarrow C_{2}-C_{3}$ લેતા:
$\Delta=\left|\begin{array}{ccc}0 & 0 & 1 \\ a-b & b-c & c \\ a^{3}-b^{3} & b^{3}-c^{3} & c^{3}\end{array}\right|$.
$C_{1}$ માંથી $(a-b)$ અને $C_{2}$ માંથી $(b-c)$ સામાન્ય લેતા:
$\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 1 & 1 & c \\ a^{2}+ab+b^{2} & b^{2}+bc+c^{2} & c^{3}\end{array}\right|$.
$C_{1} \rightarrow C_{1}-C_{2}$ લેતા:
$\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & c \\ a^{2}-c^{2}+ab-bc & b^{2}+bc+c^{2} & c^{3}\end{array}\right|$.
$\Delta=(a-b)(b-c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & c \\ (a-c)(a+c)+b(a-c) & b^{2}+bc+c^{2} & c^{3}\end{array}\right|$.
$\Delta=(a-b)(b-c)(a-c)\left|\begin{array}{ccc}0 & 0 & 1 \\ 0 & 1 & c \\ a+b+c & b^{2}+bc+c^{2} & c^{3}\end{array}\right|$.
$R_{1}$ ની સાપેક્ષ વિસ્તરણ કરતા:
$\Delta=(a-b)(b-c)(a-c)(1)[0 - (a+b+c)] = -(a-b)(b-c)(a-c)(a+b+c)$.
કારણ કે $-(a-c) = (c-a)$,તેથી:
$\Delta=(a-b)(b-c)(c-a)(a+b+c)$.
આમ,પરિણામ સાબિત થાય છે.
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