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Question Number 42009 by 737954 last updated on 16/Aug/18
if (a+bω+cω^2 )+(aω+bω^2 +c)^2 +(aω^2 +b+cω)^2 =0 then prove that a=c or a+c=2b
$$\mathrm{if}\:\left(\mathrm{a}+\mathrm{b}\omega+\mathrm{c}\omega^{\mathrm{2}} \right)+\left(\mathrm{a}\omega+\mathrm{b}\omega^{\mathrm{2}} +\mathrm{c}\right)^{\mathrm{2}} +\left(\mathrm{a}\omega^{\mathrm{2}} +\mathrm{b}+\mathrm{c}\omega\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\mathrm{then}\:\mathrm{prove}\:\mathrm{that}\:\mathrm{a}=\mathrm{c}\:\:\mathrm{or}\:\:\:\mathrm{a}+\mathrm{c}=\mathrm{2b} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Aug/18
a^2 +a^2 w^2 +a^2 w^4 +b^2 w^2 +b^2 w^4 +b^2 +c^2 w^4 +c^2 +c^2 w^2 +2abw+2abw^3 +2abw^2 +2bcw^3 +2bcw^2 +2bcw +2acw^2 +2acw+2acw^3 =a^2 (1+w^2 +w)+b^2 (w^2 +w+1)+c^2 (w+1+w^2 )+ 2ab(w+w^3 +w^2 )+2bc(w^3 +w^2 +w)+2ac(w^2 +w+w^3 ) a×0+b×0+c×0+(1+w+w^2 )(2ab+2bc+2ac) =0×(2ab+2bc+2ac =0 pls check the question because given expression=0 on simplification but can not prove a=c ora+c=2b w^4 =w^3 .w=1.w=w 1+w+w^2 =0 formula i have squared and then added identical terms by mind calculation becauze (A+B+C)^2 =A^2 +B^ +C^2 +2AB+2BC+2AC
$${a}^{\mathrm{2}} +{a}^{\mathrm{2}} {w}^{\mathrm{2}} +{a}^{\mathrm{2}} {w}^{\mathrm{4}} +{b}^{\mathrm{2}} {w}^{\mathrm{2}} +{b}^{\mathrm{2}} {w}^{\mathrm{4}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} {w}^{\mathrm{4}} +{c}^{\mathrm{2}} +{c}^{\mathrm{2}} {w}^{\mathrm{2}} \\ $$$$+\mathrm{2}{abw}+\mathrm{2}{abw}^{\mathrm{3}} +\mathrm{2}{abw}^{\mathrm{2}} +\mathrm{2}{bcw}^{\mathrm{3}} +\mathrm{2}{bcw}^{\mathrm{2}} +\mathrm{2}{bcw} \\ $$$$+\mathrm{2}{acw}^{\mathrm{2}} +\mathrm{2}{acw}+\mathrm{2}{acw}^{\mathrm{3}} \\ $$$$={a}^{\mathrm{2}} \left(\mathrm{1}+{w}^{\mathrm{2}} +{w}\right)+{b}^{\mathrm{2}} \left({w}^{\mathrm{2}} +{w}+\mathrm{1}\right)+{c}^{\mathrm{2}} \left({w}+\mathrm{1}+{w}^{\mathrm{2}} \right)+ \\ $$$$\mathrm{2}{ab}\left({w}+{w}^{\mathrm{3}} +{w}^{\mathrm{2}} \right)+\mathrm{2}{bc}\left({w}^{\mathrm{3}} +{w}^{\mathrm{2}} +{w}\right)+\mathrm{2}{ac}\left({w}^{\mathrm{2}} +{w}+{w}^{\mathrm{3}} \right) \\ $$$${a}×\mathrm{0}+{b}×\mathrm{0}+{c}×\mathrm{0}+\left(\mathrm{1}+{w}+{w}^{\mathrm{2}} \right)\left(\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ac}\right) \\ $$$$=\mathrm{0}×\left(\mathrm{2}{ab}+\mathrm{2}{bc}+\mathrm{2}{ac}\right. \\ $$$$=\mathrm{0} \\ $$$$\:{pls}\:{check}\:{the}\:{question} \\ $$$${because}\:{given}\:{expression}=\mathrm{0}\:{on}\:{simplification} \\ $$$${but}\:{can}\:{not}\:{prove}\:{a}={c}\:{ora}+{c}=\mathrm{2}{b} \\ $$$$ \\ $$$${w}^{\mathrm{4}} ={w}^{\mathrm{3}} .{w}=\mathrm{1}.{w}={w} \\ $$$$\mathrm{1}+{w}+{w}^{\mathrm{2}} =\mathrm{0}\:\:\:{formula} \\ $$$$ \\ $$$${i}\:{have}\:{squared}\:{and}\:{then}\:{added}\:{identical}\:{terms} \\ $$$${by}\:{mind}\:{calculation} \\ $$$${becauze} \\ $$$$\left({A}+{B}+{C}\right)^{\mathrm{2}} ={A}^{\mathrm{2}} +{B}^{} +{C}^{\mathrm{2}} +\mathrm{2}{AB}+\mathrm{2}{BC}+\mathrm{2}{AC} \\ $$$$ \\ $$

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