Question
Using properties of scalar triple product, prove that
$\left[\begin{array}{lll}\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}\end{array}\right]=2\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
$\left[\begin{array}{lll}\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}\end{array}\right]=2\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$
$\begin{array}{r}=\bar{a} \cdot\{(\vec{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{a})\}+ \\ \quad \bar{b} \cdot\{(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{a})\} \\ =\bar{a} \cdot(\bar{b} \times \bar{c})+\bar{a} \cdot(\bar{b} \times \bar{a})+\bar{a} \cdot(\bar{c} \times \bar{a})+ \\ \bar{b} \cdot(\bar{b} \times \bar{c})+\bar{b} \cdot(\bar{b} \times \bar{a})+\bar{b} \cdot(\bar{c} \times \bar{a})\end{array}$
$\begin{aligned} & =[\bar{a} \bar{b} \bar{c}]+[\bar{a} \bar{b} \bar{a}]+[\bar{a} \bar{c} \bar{a}]+ \\ & {[\bar{b} \bar{b} \bar{c}]+[\bar{b} \bar{b} \bar{a}]+[\bar{b} \bar{c} \bar{a}]} \\ & \end{aligned}$
$\begin{aligned} & =[\bar{a} \bar{b} \bar{c}]+0+0+0+0+[\bar{a} \bar{b} \bar{c}] \\ & =2[\bar{a} \bar{b} \bar{c}] \\ & =\text { RHS. }\end{aligned}$
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