If $[\bar{a} \bar{b} \bar{c}] \neq 0$, then $\frac{[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad…

If $[\bar{a} \bar{b} \bar{c}] \neq 0$, then $\frac{[\bar{a}+\bar{b} \quad \bar{b}+\bar{c} \quad \bar{c}+\bar{a}]}{[\bar{b} \bar{c} \bar{a}]}=$
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Solution

$\left[\begin{array}{lll}\bar{a}+\bar{b} & \bar{b}+\bar{c} & \bar{c}+\bar{a}\end{array}\right]$ $=(\bar{a}+\bar{b}) \cdot[(\bar{b}+\bar{c}) \times(\bar{c}+\bar{a})]$ $=(\bar{a}+\bar{b}) \cdot[(\bar{b} \times \bar{c})+(\bar{b} \times \bar{a})+(\bar{c} \times \bar{c})+(\bar{c} \times \bar{a})]$ $=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{a} \cdot(\bar{b} \times \bar{a})]+[\bar{a} \cdot(\bar{c} \times \bar{a})]+[\bar{b} \cdot(\bar{b} \times \bar{c})]+[\bar{b} \cdot(\bar{b} \times \bar{a})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]$ $=[\bar{a} \cdot(\bar{b} \times \bar{c})]+[\bar{b} \cdot(\bar{c} \times \bar{a})]$ $=2[\bar{a} \cdot(\bar{b} \times \bar{c})] \quad=2\left[\begin{array}{lll}\bar{a} & \bar{b} & \bar{c}\end{array}\right]$ Hence give expression $=\frac{2[\bar{a} \bar{b} \bar{c}]}{[\bar{a} \bar{b} \bar{c}]}=2$

Asked in: MHT CET 2020 (16 Oct Shift 1)

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