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Question

The value of determinant $$\begin{vmatrix} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix}$$ is


A
a3+b3+c33abc
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B
(a3+b3+c33abc)
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C
(a3+b3+c3)
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D
None of these
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Solution

The correct option is B $$-\left( { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }-3abc \right) $$
Let $$\displaystyle \Delta =\begin{vmatrix} b+c & a-b & a \\ c+a & b-c & b \\ a+b & c-a & c \end{vmatrix}$$

Applying $${ C }_{ 1 }\rightarrow { C }_{ 1 }+{ C }_{ 3 },{ C }_{ 2 }\rightarrow { C }_{ 2 }-{ C }_{ 3 },$$ we get
$$\Delta =\begin{vmatrix} a+b+c & -b & a \\ a+b+c & -c & b \\ a+b+c & -a & c \end{vmatrix}=-\left( a+b+c \right) \begin{vmatrix} 1 & b & a \\ 1 & c & b \\ 1 & a & c \end{vmatrix}$$

Applying $${ R }_{ 2 }\rightarrow { R }_{ 2 }-{ R }_{ 1 },{ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 1 },$$ we get
$$\Delta =-\left( a+b+c \right) \begin{vmatrix} 1 & b & a \\ 0 & c-b & b-a \\ 0 & a-b & c-a \end{vmatrix}\\ =-\left( a+b+c \right) \left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right) \\ =-\left( { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }-3abc \right) $$

Mathematics

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