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# By using properties of determinants, show that determinant 1 1 1 a b c a^3 b^3 c^3 = (a-b)(b-c)(c-a)(a+b+c) Ex 4.2 Q : 8 (ii)

Q : 8        By using properties of determinants, show that:

$\dpi{100} (ii) \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix}=(a-b)(b-c)(c-a)(a+b+c)$

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Given determinant :

$\dpi{100} \begin{vmatrix} 1 & 1 & 1\\ a & b & c\\ a^3 &b^3 &c^3 \end{vmatrix}$,

Applying column transformation $\dpi{100} C_{1} \rightarrow C_{1}-C_{3}$ and then $\dpi{100} C_{2} \rightarrow C_{2}-C_{3}$

We get,

$\dpi{100} \triangle =\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ a^3-c^3 &b^3-c^3 & c^3 \end{vmatrix}$

$\dpi{100} =\begin{vmatrix} 0 & 0 & 1\\ a-c& b-c & c \\ (a-c)(a^2+ac+c^2) &(b-c)(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 1& 1 & c \\ (a^2+ac+c^2) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

Now, applying column transformation $\dpi{100} C_{1} \rightarrow C_{1} - C_{2}$, we have:

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a^2-b^2+ac-bc) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)\begin{vmatrix} 0 & 0 & 1\\ 0& 1 & c \\ (a-b)(a+b+c) &(b^2+bc+c^2) & c^3 \end{vmatrix}$

$\dpi{100} =(a-c)(b-c)(a-b)(a+b+c)\begin{vmatrix} 0&1 \\ 1& c \end{vmatrix}$

$\dpi{100} =-(a-c)(b-c)(a-b)(a+b+c) = (a-b)(b-c)(c-a)(a+b+c)$

Hence proved.

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