\hspace{-16}$If $\mathbf{a\;,b\;c\in\mathbb{R}}$ and $\mathbf{a+b+c=0\;,a^2+b^2+c^2=1}$\\\\ Then $\mathbf{a^4+b^4+c^4=}$
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1\begin{align*} & a^4+b^4+c^4=(a^2+b^2+c^2)^2-2(a^2b^2+.....)\\ & 1-2(a^2b^2+b^2c^2...) ...(1)\\ & now,\\ & (a+b+c)^2-2(a+bc+ca) = 1\\ & \implies (ab+bc+ac)=-\frac{1}{2}\\ & s.o.b.s\\ &\implies \Sigma a^2b^2=\frac{1}{4} \\ &\textup{plugging in (1) gives }\boxed{a^4+b^4+c^4=\frac{1}{2}} \end{align*}
