Question Number 48993 by Raj Singh last updated on 01/Dec/18
Answered by $@ty@m last updated on 01/Dec/18
![In order to save time, I′ll write a instead of a^→ LHS=[a×b b×c c×a] =(a×b).{(b×c)×(c×a)} =(a×b).{d×(c×a)} where d=b×c =(a×b).{(d.a)c−(d.c)a} =(a×b).{[b c a]c−[b c c]a} =(a×b).{[b c a]c−0} ∵[b c c]=0 =[b c a](a×b.c) =[b c a][a b c] =[a b c][a b c] =[a b c]^2 =RHS](https://www.tinkutara.com/question/Q48997.png)
$${In}\:{order}\:{to}\:{save}\:{time}, \\ $$$${I}'{ll}\:{write}\:\mathrm{a}\:{instead}\:{of}\:\overset{\rightarrow} {\mathrm{a}} \\ $$$${LHS}=\left[{a}×{b}\:{b}×{c}\:{c}×{a}\right] \\ $$$$=\left({a}×{b}\right).\left\{\left({b}×{c}\right)×\left({c}×{a}\right)\right\} \\ $$$$=\left({a}×{b}\right).\left\{{d}×\left({c}×{a}\right)\right\}\:{where}\:{d}={b}×{c} \\ $$$$=\left({a}×{b}\right).\left\{\left({d}.{a}\right){c}−\left({d}.{c}\right){a}\right\} \\ $$$$=\left({a}×{b}\right).\left\{\left[{b}\:{c}\:{a}\right]{c}−\left[{b}\:{c}\:{c}\right]{a}\right\} \\ $$$$=\left({a}×{b}\right).\left\{\left[{b}\:{c}\:{a}\right]{c}−\mathrm{0}\right\}\:\:\because\left[{b}\:{c}\:{c}\right]=\mathrm{0} \\ $$$$=\left[{b}\:{c}\:{a}\right]\left({a}×{b}.{c}\right) \\ $$$$=\left[{b}\:{c}\:{a}\right]\left[{a}\:{b}\:{c}\right] \\ $$$$=\left[{a}\:{b}\:{c}\right]\left[{a}\:{b}\:{c}\right] \\ $$$$=\left[{a}\:{b}\:{c}\right]^{\mathrm{2}} ={RHS} \\ $$