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If a+b+c = 0 , a^{2}+b^{2}+c^{2}=1 , find [ a^{4}+b^{4}+c^{4}.

Note by Shubham Bagrecha 4 years, 3 months ago

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See \(a+b+c\)=0 So \(a^2+b^2+c^2+2(a+b+c)\)=0, or

\(ab+bc+ca\)= \(-0.5\)

Thus \(a^4+b^4+c^4\)

=\([a^2+b^2+c^2]^2-2[a^2b^2+b^2c^2+c^2a^2]\)

=\(1-2((ab+bc+ca)^2-2(ab^2c+abc^2+a^2bc))\)

=\(1-2((-0.5)^2-2abc(a+b+c))\)

=\(1-2(0.25)\)

=\(0.5\) – Shourya Pandey · 4 years, 3 months ago

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This question looks like,the brilliant question – Beakal Tiliksew · 4 years, 3 months ago

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## Comments

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TopNewestSee \(a+b+c\)=0 So \(a^2+b^2+c^2+2(a+b+c)\)=0, or

\(ab+bc+ca\)= \(-0.5\)

Thus \(a^4+b^4+c^4\)

=\([a^2+b^2+c^2]^2-2[a^2b^2+b^2c^2+c^2a^2]\)

=\(1-2((ab+bc+ca)^2-2(ab^2c+abc^2+a^2bc))\)

=\(1-2((-0.5)^2-2abc(a+b+c))\)

=\(1-2(0.25)\)

=\(0.5\) – Shourya Pandey · 4 years, 3 months ago

Log in to reply

This question looks like,the brilliant question – Beakal Tiliksew · 4 years, 3 months ago

Log in to reply