Question-59060

Question Number 59060 by Tawa1 last updated on 04/May/19

Answered by tanmay last updated on 04/May/19

a^3 +b^3 +c^3 −3abc (a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca) (a+b+c)(a^2 +b^2 +c^2 +abw+abw^2 +bcw+bcw^2 +caw+caw^2 ) we know w^3 =1 and 1+w+w^2 =0 (a+b+c)(a^2 +abw+acw^2 +b^2 w^3 +abw^2 +bcw^4 +c^2 w^3 +bcw^2 +acw) (a+b+c){a(a+bw+cw^2 )+bw^2 (a+bw+cw^2 )+cw(a+bw+cw^2 )} (a+b+c)(a+bw+cw^2 )(a+bw^2 +cw)

$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc} \\ $$$$\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right) \\ $$$$\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{abw}+{abw}^{\mathrm{2}} +{bcw}+{bcw}^{\mathrm{2}} +{caw}+{caw}^{\mathrm{2}} \right) \\ $$$${we}\:{know}\:{w}^{\mathrm{3}} =\mathrm{1}\:\:{and}\:\mathrm{1}+{w}+{w}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{abw}+{acw}^{\mathrm{2}} +{b}^{\mathrm{2}} {w}^{\mathrm{3}} +{abw}^{\mathrm{2}} +{bcw}^{\mathrm{4}} +{c}^{\mathrm{2}} {w}^{\mathrm{3}} +{bcw}^{\mathrm{2}} +{acw}\right) \\ $$$$\left({a}+{b}+{c}\right)\left\{{a}\left({a}+{bw}+{cw}^{\mathrm{2}} \right)+{bw}^{\mathrm{2}} \left({a}+{bw}+{cw}^{\mathrm{2}} \right)+{cw}\left({a}+{bw}+{cw}^{\mathrm{2}} \right)\right\} \\ $$$$\left({a}+{b}+{c}\right)\left({a}+{bw}+{cw}^{\mathrm{2}} \right)\left({a}+{bw}^{\mathrm{2}} +{cw}\right) \\ $$

Commented by Tawa1 last updated on 04/May/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by math1967 last updated on 04/May/19

(a+b+c)(a^2 +abω^2 +caω+abω+b^2 ω^3 +bcω^2 +caω^2 +bcω^3 ω+ω^3 c^2 ) =(a+b+c){a^2 +b^2 +c^2 +ab(ω^2 +ω) +bc(ω^2 +ω)+ca(ω+ω^2 )} =(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)★ =a^3 +b^3 +c^3 −3abc ★ω+ω^2 +1=0∴ω+ω^2 =−1 ω^3 =1

$$\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{ab}\omega^{\mathrm{2}} +{ca}\omega+{ab}\omega+{b}^{\mathrm{2}} \omega^{\mathrm{3}} \right. \\ $$$$\left.\:\:\:\:\:\:+{bc}\omega^{\mathrm{2}} +{ca}\omega^{\mathrm{2}} +{bc}\omega^{\mathrm{3}} \omega+\omega^{\mathrm{3}} {c}^{\mathrm{2}} \right) \\ $$$$=\left({a}+{b}+{c}\right)\left\{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ab}\left(\omega^{\mathrm{2}} +\omega\right)\right. \\ $$$$\left.\:\:\:+{bc}\left(\omega^{\mathrm{2}} +\omega\right)+{ca}\left(\omega+\omega^{\mathrm{2}} \right)\right\} \\ $$$$=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)\bigstar \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} −\mathrm{3}{abc} \\ $$$$\bigstar\omega+\omega^{\mathrm{2}} +\mathrm{1}=\mathrm{0}\therefore\omega+\omega^{\mathrm{2}} =−\mathrm{1} \\ $$$$\:\:\:\:\:\omega^{\mathrm{3}} =\mathrm{1} \\ $$

Commented by Tawa1 last updated on 04/May/19

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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