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(B) Given $a+b+c=0$. We know that $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) = 0$. Thus,$q + 2(ab+bc+ca) = 0$,which implies $ab+bc+ca = -\frac{q}{2}$. Squ

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$q^2 = 2r$ always

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(B) Given $a+b+c=0$. We know that $(a+b+c)^2 = a^2+b^2+c^2 + 2(ab+bc+ca) = 0$. Thus,$q + 2(ab+bc+ca) = 0$,which implies $ab+bc+ca = -\frac{q}{2}$. Squaring both sides,$(ab+bc+ca)^2 = \frac{q^2}{4}$. Also,$(ab+bc+ca)^2 = a^2b^2+b^2c^2+c^2a^2 + 2abc(a+b+c) = a^2b^2+b^2c^2+c^2a^2 + 0 = \frac{q^2}{4}$. Now,consider $q^2 = (a^2+b^2+c^2)^2 = a^4+b^4+c^4 + 2(a^2b^2+b^2c^2+c^2a^2)$. Substituting $r = a^4+b^4+c^4$ and $a^2b^2+b^2c^2+c^2a^2 = \frac{q^2}{4}$,we get: $q^2 = r + 2(\frac{q^2}{4}) = r + \frac{q^2}{2}$. Rearranging the terms,$q^2 - \frac{q^2}{2} = r$,which simplifies to $\frac{q^2}{2} = r$,or $q^2 = 2r$.

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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