If #a +b+c =0# and #a^2+b^2+c^2 =1# , then the value of#a^4+b^4+c^4# is?

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Answer 1 · 2017-03-11T12:10:16

#1/2#

#(a+b+c)^2=a^2+b^2+c^2+2(ab+ac+bc)=0# so

#ab+ac+bc=-1/2#

#(a^2+b^2+c^2)^2=a^4+b^4+c^4+2(a^ 2b^2+a^2c^2+b^2c^2)=1# so

#a^4+b^4+c^4=1-2(a^ 2b^2+a^2c^2+b^2c^2)# but

#(ab+ac+bc)^2=a^ 2b^2+a^2c^2+b^2c^2+2(a^2bc+ab^2c+abc^2) = 1/4#

and consequently

#a^ 2b^2+a^2c^2+b^2c^2 = 1/4-2(a+b+c)abc = 1/4#

Finally

#a^4+b^4+c^4=1-2/4=1/2#