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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 5 years, 2 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 \(a^4+b^4+c^4\)

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

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

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

=\(1-2(0.25)\)

=\(0.5\)

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

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

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

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

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

=\(1-2(0.25)\)

=\(0.5\)

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This question looks like,the brilliant question

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