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Question-57618




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
∣b+c    a−b    a∣  ∣c+a     b−c    b∣  ∣a+b     c−a    c∣  ∣b       a      a∣+∣b   −b    a∣+∣c    a    a∣+∣c    −b  a ∣  ∣c       b      b ∣     ∣c     −c  b∣    ∣a   b     b∣    ∣a  −c     b∣  ∣a        c     c∣     ∣a     −a  c∣   ∣b    c     c∣    ∣b     −a   c∣  now △=△_1 +△_2 +△_3 +△_4   △_1 =△_2 =△_3 =0   reason collumn identical  △=△_4   =(−1)∣c    b     a∣                  ∣a      c    b∣                  ∣b      a     c∣  =(−1)×{c(c^2 −ab)−b(ac−b^2 )+a(a^2 −bc)}  =(−1)(c^3 −abc−abc+b^3 +a^3 −abc)  =3abc−a^3 −b^3 −c^3
$$\mid{b}+{c}\:\:\:\:{a}−{b}\:\:\:\:{a}\mid \\ $$$$\mid{c}+{a}\:\:\:\:\:{b}−{c}\:\:\:\:{b}\mid \\ $$$$\mid{a}+{b}\:\:\:\:\:{c}−{a}\:\:\:\:{c}\mid \\ $$$$\mid{b}\:\:\:\:\:\:\:{a}\:\:\:\:\:\:{a}\mid+\mid{b}\:\:\:−{b}\:\:\:\:{a}\mid+\mid{c}\:\:\:\:{a}\:\:\:\:{a}\mid+\mid{c}\:\:\:\:−{b}\:\:{a}\:\mid \\ $$$$\mid{c}\:\:\:\:\:\:\:{b}\:\:\:\:\:\:{b}\:\mid\:\:\:\:\:\mid{c}\:\:\:\:\:−{c}\:\:{b}\mid\:\:\:\:\mid{a}\:\:\:{b}\:\:\:\:\:{b}\mid\:\:\:\:\mid{a}\:\:−{c}\:\:\:\:\:{b}\mid \\ $$$$\mid{a}\:\:\:\:\:\:\:\:{c}\:\:\:\:\:{c}\mid\:\:\:\:\:\mid{a}\:\:\:\:\:−{a}\:\:{c}\mid\:\:\:\mid{b}\:\:\:\:{c}\:\:\:\:\:{c}\mid\:\:\:\:\mid{b}\:\:\:\:\:−{a}\:\:\:{c}\mid \\ $$$${now}\:\bigtriangleup=\bigtriangleup_{\mathrm{1}} +\bigtriangleup_{\mathrm{2}} +\bigtriangleup_{\mathrm{3}} +\bigtriangleup_{\mathrm{4}} \\ $$$$\bigtriangleup_{\mathrm{1}} =\bigtriangleup_{\mathrm{2}} =\bigtriangleup_{\mathrm{3}} =\mathrm{0}\:\:\:{reason}\:{collumn}\:{identical} \\ $$$$\bigtriangleup=\bigtriangleup_{\mathrm{4}} \\ $$$$=\left(−\mathrm{1}\right)\mid{c}\:\:\:\:{b}\:\:\:\:\:{a}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid{a}\:\:\:\:\:\:{c}\:\:\:\:{b}\mid \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mid{b}\:\:\:\:\:\:{a}\:\:\:\:\:{c}\mid \\ $$$$=\left(−\mathrm{1}\right)×\left\{{c}\left({c}^{\mathrm{2}} −{ab}\right)−{b}\left({ac}−{b}^{\mathrm{2}} \right)+{a}\left({a}^{\mathrm{2}} −{bc}\right)\right\} \\ $$$$=\left(−\mathrm{1}\right)\left({c}^{\mathrm{3}} −{abc}−{abc}+{b}^{\mathrm{3}} +{a}^{\mathrm{3}} −{abc}\right) \\ $$$$=\mathrm{3}{abc}−{a}^{\mathrm{3}} −{b}^{\mathrm{3}} −{c}^{\mathrm{3}} \\ $$

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