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Prove that ∣ ∣ ∣ ∣ ∣ a 2 2 a B B 2 B 2 a 2 2 a B 2 a B B 2 a 2 ∣ ∣ ∣ ∣ ∣ = ( a 3 + B 3 ) 2

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प्रश्न

Prove that

\[\begin{vmatrix}a^2 & 2ab & b^2 \\ b^2 & a^2 & 2ab \\ 2ab & b^2 & a^2\end{vmatrix} = \left( a^3 + b^3 \right)^2\]
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उत्तर

\[\text{ Let LHS }= ∆ = \begin{vmatrix} a^2 & 2ab & b^2 \\ b^2 & a^2 & 2ab \\2ab & b^2 & a^2 \end{vmatrix}\] 
\[ = a^2 \begin{vmatrix} a^2 & 2ab \\ b^2 & a^2 \end{vmatrix} - \left( 2ab \right) \begin{vmatrix} b^2 & 2ab \\2ab & a^2 \end{vmatrix} + b^2 \begin{vmatrix} b^2 & a^2 \\2ab & b^2 \end{vmatrix} \left[\text{ Expanding }\right]\] 
\[ = a^2 \left( a^4 - 2a b^3 \right) - \left( 2ab \right)\left( b^2 a^2 - 4 a^2 b^2 \right) + b^2 \left( b^4 - 2 a^3 b \right)\] 
\[ = a^6 - 2 a^3 b^3 - 2 a^3 b^3 + 8 a^3 b^3 + b^6 - 2 a^3 b^3 \] 
\[ = a^6 + 2 a^3 b^3 + b^6 \] 
\[ = \left( a^3 \right)^2 + 2 a^3 b^3 + \left( b^3 \right)^2 \] 
\[ = \left( a^3 + b^3 \right)^2 \] 
\[ = RHS\]
Hence proved.

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अध्याय 5: Determinants - Exercise 6.2 [पृष्ठ ६०]

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आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
अध्याय 5 Determinants
Exercise 6.2 | Q 28 | पृष्ठ ६०

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