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Question Number 177331 by mr W last updated on 03/Oct/22

$${a}+{b}+{c}=\mathrm{0} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{4} \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =? \\ $$

Answered by a.lgnaoui last updated on 04/Oct/22

a^2 +b^2 +c^2 =(a+b+c)^2 −2(ab+bc+ac)=4 ab+bc+ac=−2 (1)( a^4 +b^4 +c^4 =(a^2 +b^2 +c^2 )^2 −2[(ab)^2 +(bc)^2 +(ac)^2 ] a^4 +b^4 +c^4 =4^2 −2[(ab+bc+ac)^2 −(ab×bc+ab×ac+bc×ab+bc×ac+ac×bc+ac×ab)= 4^2 −2[(ab+bc+ac)^2 −2(ab×bc+ab×ac+bcac)] 4^2 −2[(ab+bc+ac)^2 −2abc(a+b+c)] =4^2 −2[(−2)^2 −0]=8 donc a^4 +b^4 +c^4 =8

$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ac}\right)=\mathrm{4} \\ $$$${ab}+{bc}+{ac}=−\mathrm{2}\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{1}\right)\left(\right. \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{2}\left[\left({ab}\right)^{\mathrm{2}} +\left({bc}\right)^{\mathrm{2}} +\left({ac}\right)^{\mathrm{2}} \right] \\ $$$${a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{4}^{\mathrm{2}} −\mathrm{2}\left[\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} −\left({ab}×{bc}+{ab}×{ac}+{bc}×{ab}+{bc}×{ac}+{ac}×{bc}+{ac}×{ab}\right)=\right. \\ $$$$\mathrm{4}^{\mathrm{2}} −\mathrm{2}\left[\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}×{bc}+{ab}×{ac}+{bcac}\right)\right] \\ $$$$\mathrm{4}^{\mathrm{2}} −\mathrm{2}\left[\left({ab}+{bc}+{ac}\right)^{\mathrm{2}} −\mathrm{2}{abc}\left({a}+{b}+{c}\right)\right] \\ $$$$=\mathrm{4}^{\mathrm{2}} −\mathrm{2}\left[\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{0}\right]=\mathrm{8} \\ $$$${donc}\:\:\:\:{a}^{\mathrm{4}} +{b}^{\mathrm{4}} +{c}^{\mathrm{4}} =\mathrm{8} \\ $$