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$[\mathbf{b} \times \mathbf{c}, \mathbf{c} \times \mathbf{a}, \mathbf{a} \times \mathbf{b}]$ is equal to
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Verified Answer
The correct answer is:
$[\mathbf{a} \mathbf{b} \mathbf{c}]^2$
$\begin{aligned}
& {[\mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]=(\mathbf{b} \times \mathbf{c}) .[(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})]} \\
& \text { Let } \mathbf{a} \times \mathbf{b}=\mathbf{d} \\
& \text { so, }(\mathbf{b} \times \mathbf{c})[(\mathbf{c} \times \mathbf{a}) \times \mathbf{d}]=(\mathbf{b} \times \mathbf{c})[(\mathbf{d} . \mathbf{a}) \mathbf{c}-(\mathbf{d} . \mathbf{c}) \cdot \mathbf{a}] \\
& =(\mathbf{b} \times \mathbf{c})[\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}-(\mathbf{a} \times \mathbf{b}) \mathbf{c} \cdot \mathbf{a}] \\
& =(\mathbf{b} \times \mathbf{c})[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{a}=\mathbf{a} \cdot[\mathbf{b} \times \mathbf{c}] \cdot[\mathbf{a} \mathbf{b} \mathbf{c}] \\
& =[\mathbf{a} \mathbf{b} \mathbf{c}][\mathbf{a} \mathbf{b} \mathbf{c}]=[\mathbf{a} \mathbf{b} \mathbf{c}]^2
\end{aligned}$
& {[\mathbf{b} \times \mathbf{c} \mathbf{c} \times \mathbf{a} \mathbf{a} \times \mathbf{b}]=(\mathbf{b} \times \mathbf{c}) .[(\mathbf{c} \times \mathbf{a}) \times(\mathbf{a} \times \mathbf{b})]} \\
& \text { Let } \mathbf{a} \times \mathbf{b}=\mathbf{d} \\
& \text { so, }(\mathbf{b} \times \mathbf{c})[(\mathbf{c} \times \mathbf{a}) \times \mathbf{d}]=(\mathbf{b} \times \mathbf{c})[(\mathbf{d} . \mathbf{a}) \mathbf{c}-(\mathbf{d} . \mathbf{c}) \cdot \mathbf{a}] \\
& =(\mathbf{b} \times \mathbf{c})[\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}-(\mathbf{a} \times \mathbf{b}) \mathbf{c} \cdot \mathbf{a}] \\
& =(\mathbf{b} \times \mathbf{c})[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{a}=\mathbf{a} \cdot[\mathbf{b} \times \mathbf{c}] \cdot[\mathbf{a} \mathbf{b} \mathbf{c}] \\
& =[\mathbf{a} \mathbf{b} \mathbf{c}][\mathbf{a} \mathbf{b} \mathbf{c}]=[\mathbf{a} \mathbf{b} \mathbf{c}]^2
\end{aligned}$
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