Question

# Prove that:$$\begin{vmatrix} { \left( b+c \right) }^{ 2 } & ab & ca \\ ab & { \left( c+a \right) }^{ 2 } & bc \\ ca & bc & { \left( a+b \right) }^{ 2 } \end{vmatrix}=2abc{(a+b+c)}^{3}$$

Solution

## $$L:H:S=\left| \begin{matrix} \left( b+c \right) ^{ 2 } & ab & ca \\ ab & \left( c+a \right) ^{ 2 } & bc \\ ca & bc & \left( a+b \right) ^{ 2 } \end{matrix} \right|$$$$= \dfrac {1}{abc}\left| \begin{matrix} a\left( b+c \right) ^{ 2 } & a^{ 2 }b & a^{ 2 }c \\ ab^{ 2 } & b\left( c+a \right) ^{ 2 } & b^{ 2 }c \\ ac^{ 2 } & bc^{ 2 } & c\left( a+b \right) ^{ 2 } \end{matrix} \right|$$$$=\left| \begin{matrix} \left( b+c \right) ^{ 2 } & a^{ 2 } & a^{ 2 } \\ b^{ 2 } & \left( c+a \right) ^{ 2 } & b^{ 2 } \\ c^{ 2 } & c^{ 2 } & \left( a+b \right) ^{ 2 } \end{matrix} \right|$$$$R_{1}+ C_{1}- C_{3},C_{2}- C_{2}- C_{3}$$$$= \left| \begin{matrix} (b+c-a)(b+c+a) & 0 & a^{ 2 } \\ 0 & \left( c+a+b \right) (c+a-b) & b^{ 2 } \\ (c-a-b)(c+a+b) & \left( c-a-b \right) (c+a+b) & \left( a+b \right) ^{ 2 } \end{matrix} \right|$$$$(a+b+c) ^{2}\left| \begin{matrix} b+c-a & 0 & a^{ 2 } \\ 0 & c+a-b & b^{ 2 } \\ c-a-b & c-a-b & \left( a+b \right) ^{ 2 } \end{matrix} \right|$$$$R_{3}+R_{3}-(R_{1}+R_{2})$$$$(a+b+c)^{2}\left| \begin{matrix} b+c-a & 0 & a^{ 2 } \\ 0 & c+a-b & b^{ 2 } \\ -2b & -2a & 2ab \end{matrix} \right|$$$$\dfrac {(a+b+c)^{2}}{ab}\left| \begin{matrix} a(b+c-a) & 0 & a^{ 2 } \\ 0 & b(c+a-b) & b^{ 2 } \\ -2ab & -2ab & 2ab \end{matrix} \right|$$$$\dfrac {(a+b+c)^{2}}{ab}\left| \begin{matrix} a+b+c & a^{ 2 } & a^{ 2 } \\ b^{ 2 } & bc+ab & b^{ 2 } \\ U & U & 2ab \end{matrix} \right|$$$$2abc(a+b+c)^{3}$$$$R:H:S$$ (Proved) Mathematics

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