Question
Using properties of scalar triple product, prove that $\left[\begin{array}{llll}\overline{ a }+\overline{ b } & \overline{ b }+\overline{ c } & \overline{ c }+\overline{ a }\end{array}\right]=2\left[\begin{array}{lll}\overline{ a } & \overline{ b } & \overline{ c }\end{array}\right]$.

Answer

$\text { L.H.S }=\left[\begin{array}{lll}\overline{ a }+\overline{ b } & \overline{ b }+\overline{ c } & \overline{ c }+\overline{ a }\right]$
$=(\overline{ a }+\overline{ b }) \cdot[(\overline{ b }+\overline{ c }) \times(\overline{ c }+\overline{ a })]$
$=(\overline{ a }+\overline{ b }) \cdot[\overline{ b } \times \overline{ c }+\overline{ b } \times \overline{ a }+\overline{ c } \times \overline{ c }+\overline{ c } \times \overline{ a }]$
$=(\overline{ a }+\overline{ b }) \cdot[\overline{ b } \times \overline{ c }+\overline{ b } \times \overline{ a }+\overline{ c } \times \overline{ a }] \quad \ldots[\because \overline{ c } \times \overline{ c }=\overline{0}]$
$=\overline{ a } \cdot[(\overline{ b } \times \overline{ c })+(\overline{ b } \times \overline{ a })+(\overline{ c } \times \overline{ a })]+\overline{ b } \cdot[(\overline{ b } \times \overline{ c })+(\overline{ b } \times \overline{ a })+(\overline{ c } \times \overline{ a })]$
$=\overline{ a } \cdot(\overline{ b } \times \overline{ c })+\overline{ a } \cdot(\overline{ b } \times \overline{ a })+\overline{ a } \cdot(\overline{ c } \times \overline{ a })+\overline{ b } \cdot(\overline{ b } \times \overline{ c })+\overline{ b }(\overline{ b } \times \overline{ a })+\overline{ b }(\overline{ c } \times \overline{ a })$
$=[\overline{ a } \overline{ b } \overline{ c }]+[\overline{ a } \overline{ b } \overline{ a }]+[\overline{ a } \overline{ c } \overline{ a }]+[\overline{ b } \overline{ b } \overline{ c }]+[\overline{ b } \overline{ b } \overline{ a }]+[\overline{ b } \overline{ c } \overline{ a }]$
$=[\overline{ a } \overline{ b } \overline{ c }]+0+0+0+0+[\overline{ a } \overline{ b } \overline{ c }]$
$=2[\overline{ a } \overline{ b } \overline{ c }]$
$=R \cdot H \cdot S$

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