Question

prove that value LHS=RHS ​

Answer 1

Answer:

which class is that

Step-by-step explanation:

please mark as brainlest answer

Answer 2

$\large\maltese \: \: \: \sf \underline{ \underline{Question \: :}}$

• Prove that,

$\begin{array}{ |ccc| } \bf b + c& \bf a& \bf a \\ \\ \bf b& \bf c + a& \bf b \\ \\ \bf c& \bf c& \bf a + b\end{array} = \bf 4abc$

$\large \maltese \: \: \: \sf \underline{ \underline{Solution\: :}}$

$\: \: \: \: \: \: \: \: \large \bf \: L \: H \: S$

$\longrightarrow \: \begin{array}{ |ccc| } \sf \: b + c& \sf a& \sf a \\ \\ \sf b& \sf c + a& \sf b \\ \\ \sf c& \sf c& \sf a + b\end{array}$

$\longrightarrow \begin{array}{| ccc| } \sf \: b + c & \sf \: a & \sf \: a \\ \\ \sf \: b + c & \sf \: a + 2c& \sf \: a + 2b \\ \\ \sf \: c& \sf \: c & \sf \: a + b\end{array} \: \: \bf \: \: \: \: \: \: \: \bigg[C_2 + C_3 \bigg]$

$\longrightarrow \begin{array}{| ccc| } \sf \: b + c& \sf \:a & \sf \: a \\ \\ \sf \: 0& \sf \:2c& \sf \:2b \\ \\ \sf \: c& \sf \: c & \sf \: a + b\end{array}\: \: \: \: \: \: \bf\bigg[C_2 - C_1 \bigg]$

$\longrightarrow \sf \: \bigg(b + c \bigg) \bigg \{(2c)(a + b) - (2b)(c) \bigg \} + 0 + c \bigg(2ab - 2ac \bigg)$

$\sf \longrightarrow \bigg(b + c \bigg) \bigg \{ 2ac + \cancel{ 2bc }- \cancel{ 2bc} \bigg \} + 2abc - 2a {c}^{2}$

$\longrightarrow \sf \: 2abc \: + \cancel{2a {c}^{2}} + 2abc - \cancel{ 2a {c}^{2} }$

$\longrightarrow { \underline{\boxed{\bf{4abc}} \: \: }}_{ \bigstar \star} \: \: \big(\bf RHS \big)\: \: \: \: \: \: \: \: \: \small{\bf{[ Proved ]}}$