If a, b and c are non-zero numbers,then Delta | vedclass

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(D) To evaluate $\Delta = \left| \begin{array}{ccc} b^2c^2 & bc & b+c \ c^2a^2 & ca & c+a \ a^2b^2 & ab & a+b \end{array} ight|$,we multiply $R_1$

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(D) To evaluate $\Delta = \left| \begin{array}{ccc} b^2c^2 & bc & b+c \ c^2a^2 & ca & c+a \ a^2b^2 & ab & a+b \end{array} ight|$,we multiply $R_1$ by $a$,$R_2$ by $b$,and $R_3$ by $c$ and divide the determinant by $abc$:$\Delta = \frac{1}{abc} \left| \begin{array}{ccc} ab^2c^2 & abc & a(b+c) \ a^2bc^2 & abc & b(c+a) \ a^2b^2c & abc & c(a+b) \end{array} ight|$Taking $abc$ common from $C_1$ and $abc$ common from $C_2$:$\Delta = \frac{abc \cdot abc}{abc} \left| \begin{array}{ccc} bc & 1 & ab+ac \ ac & 1 & bc+ab \ ab & 1 & ac+bc \end{array} ight| = abc \left| \begin{array}{ccc} bc & 1 & ab+ac \ ac & 1 & bc+ab \ ab & 1 & ac+bc \end{array} ight|$Applying $C_3 	o C_3 + C_1$ is not directly helpful here,instead observe that if we add $C_1$ to $C_3$,we get $ab+bc+ca$ in each element of $C_3$:$\Delta = abc \left| \begin{array}{ccc} bc & 1 & ab+bc+ca \ ac & 1 & ab+bc+ca \ ab & 1 & ab+bc+ca \end{array} ight|$Taking $(ab+bc+ca)$ common from $C_3$:$\Delta = abc(ab+bc+ca) \left| \begin{array}{ccc} bc & 1 & 1 \ ac & 1 & 1 \ ab & 1 & 1 \end{array} ight|$Since $C_2$ and $C_3$ are identical,the value of the determinant is $0$.

If $a, b$ and $c$ are non-zero numbers,then $\Delta = \left| \begin{array}{ccc} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \end{array} \right|$ is equal to

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