Question Number 57618 by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19
$$\mid{b}+{c}\:\:\:\:{a}−{b}\:\:\:\:{a}\mid \\ $$$$\mid{c}+{a}\:\:\:\:\:{b}−{c}\:\:\:\:{b}\mid \\ $$$$\mid{a}+{b}\:\:\:\:\:{c}−{a}\:\:\:\:{c}\mid \\ $$$$\mid{b}\:\:\:\:\:\:\:{a}\:\:\:\:\:\:{a}\mid+\mid{b}\:\:\:−{b}\:\:\:\:{a}\mid+\mid{c}\:\:\:\:{a}\:\:\:\:{a}\mid+\mid{c}\:\:\:\:−{b}\:\:{a}\:\mid \\ $$$$\mid{c}\:\:\:\:\:\:\:{b}\:\:\:\:\:\:{b}\:\mid\:\:\:\:\:\mid{c}\:\:\:\:\:−{c}\:\:{b}\mid\:\:\:\:\mid{a}\:\:\:{b}\:\:\:\:\:{b}\mid\:\:\:\:\mid{a}\:\:−{c}\:\:\:\:\:{b}\mid \\ $$$$\mid{a}\:\:\:\:\:\:\:\:{c}\:\:\:\:\:{c}\mid\:\:\:\:\:\mid{a}\:\:\:\:\:−{a}\:\:{c}\mid\:\:\:\mid{b}\:\:\:\:{c}\:\:\:\:\:{c}\mid\:\:\:\:\mid{b}\:\:\:\:\:−{a}\:\:\:{c}\mid \\ $$$${now}\:\bigtriangleup=\bigtriangleup_{\mathrm{1}} +\bigtriangleup_{\mathrm{2}} +\bigtriangleup_{\mathrm{3}} +\bigtriangleup_{\mathrm{4}} \\ $$$$\bigtriangleup_{\mathrm{1}} =\bigtriangleup_{\mathrm{2}} =\bigtriangleup_{\mathrm{3}} =\mathrm{0}\:\:\:{reason}\:{collumn}\:{identical} \\ $$$$\bigtriangleup=\bigtriangleup_{\mathrm{4}} \\ $$$$=\left(−\mathrm{1}\right)\mid{c}\:\:\:\:{b}\:\:\:\:\:{a}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid{a}\:\:\:\:\:\:{c}\:\:\:\:{b}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid{b}\:\:\:\:\:\:{a}\:\:\:\:\:{c}\mid \\ $$$$=\left(−\mathrm{1}\right)×\left\{{c}\left({c}^{\mathrm{2}} −{ab}\right)−{b}\left({ac}−{b}^{\mathrm{2}} \right)+{a}\left({a}^{\mathrm{2}} −{bc}\right)\right\} \\ $$$$=\left(−\mathrm{1}\right)\left({c}^{\mathrm{3}} −{abc}−{abc}+{b}^{\mathrm{3}} +{a}^{\mathrm{3}} −{abc}\right) \\ $$$$=\mathrm{3}{abc}−{a}^{\mathrm{3}} −{b}^{\mathrm{3}} −{c}^{\mathrm{3}} \\ $$