If a, b, c are three non-zero real numbers su | vedclass

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(B) Given that $a+b+c=0$.This implies $(a+c) = -b$.Squaring both sides,we get $(a+c)^{2} = (-b)^{2}$.$a^{2} + c^{2} + 2ac = b^{2}$.Therefore,$a^{2} +

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$2$

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(B) Given that $a+b+c=0$.This implies $(a+c) = -b$.Squaring both sides,we get $(a+c)^{2} = (-b)^{2}$.$a^{2} + c^{2} + 2ac = b^{2}$.Therefore,$a^{2} + c^{2} = b^{2} - 2ac$.Now,substitute this into the expression $\frac{a^{2}+b^{2}+c^{2}}{b^{2}-ca}$:$\frac{(a^{2}+c^{2}) + b^{2}}{b^{2}-ca} = \frac{(b^{2}-2ac) + b^{2}}{b^{2}-ca}$.$= \frac{2b^{2}-2ac}{b^{2}-ca} = \frac{2(b^{2}-ac)}{b^{2}-ac}$.Since $b^{2} 
eq ca$,we can cancel the term $(b^{2}-ac)$,which gives the result $2$.

If $a, b, c$ are three non-zero real numbers such that $a+b+c=0$ and $b^{2} \neq ca,$ then the value of $\frac{a^{2}+b^{2}+c^{2}}{b^{2}-ca}$ is

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