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Question:

Prove that $\left|\begin{array}{ccc}a^{2} & 2 a b & b^{2} \\ b^{2} & a^{2} & 2 a b \\ 2 a b & b^{2} & a^{2}\end{array}\right|=\left(a^{3}+b^{3}\right)^{2}$

Solution:

Let $\mathrm{LHS}=\Delta=\mid \begin{array}{lll}a^{2} & 2 a b & b^{2}\end{array}$

$b^{2} \quad a^{2} \quad 2 a b$

$\begin{array}{lll}2 a b & b^{2} & a^{2} \mid\end{array}$

$=a^{2} \mid a^{2} \quad 2 a b$

$b^{2} \quad a^{2}|-(2 a b)| b^{2} \quad 2 a b$

$2 a b \quad a^{2}\left|+b^{2}\right| b^{2} \quad a^{2}$

$2 a b \quad b^{2} \mid \quad[$ Expanding $]$c

$=a^{2}\left(a^{4}-2 a b^{3}\right)-(2 a b)\left(b^{2} a^{2}-4 a^{2} b^{2}\right)+b^{2}\left(b^{4}-2 a^{3} b\right)$

$=a^{6}-2 a^{3} b^{3}-2 a^{3} b^{3}+8 a^{3} b^{3}+b^{6}-2 a^{3} b^{3}$

$=a^{6}+2 a^{3} b^{3}+b^{6}$

$=\left(a^{3}\right)^{2}+2 a^{3} b^{3}+\left(b^{3}\right)^{2}$

$=\left(a^{3}+b^{3}\right)^{2}$

$=$ RHS

Hence proved.

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