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Let $a, b, c$ be non-zero real numbers such that $a+b+c=0$; let $q=a^{2}+b^{2}+c^{2}$ and $r=a^{4}+b^{4}+c^{4}$ Then
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The correct answer is:
$q 2=2 r$ always
$\begin{array}{l}a+b+c=0, \quad a, b, c \in R \neq 0 \\ a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=0 \\ q=a^{2}+b^{2}+c^{2}, & r=a^{4}+b^{4}+c^{4} \\ r=q^{2}-2\left(a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}\right) \\ r=q^{2}-2\left[(a b+b c+c a)^{2}-2 a b c(a+b+c)\right] \\ r=q^{2}-2\left(q^{2} / 4\right) \\ r=q^{2} / 2 \\ \text { ANS - B }\end{array}$
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