# Algebra!!!

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 $$a^4+b^4+c^4$$

=$$[a^2+b^2+c^2]^2-2[a^2b^2+b^2c^2+c^2a^2]$$

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

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

=$$1-2(0.25)$$

=$$0.5$$

- 5 years, 2 months ago

This question looks like,the brilliant question

- 5 years, 2 months ago