If a^{2}+b^{2}+c^{2}=ab+bc+ac, then the value | vedclass

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(B) Given the equation: $a^{2}+b^{2}+c^{2}=ab+bc+ca$.Rearranging the terms,we get: $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$.Multiplying the entire equation by $

Vedclass · Accepted Answer

$2$

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(B) Given the equation: $a^{2}+b^{2}+c^{2}=ab+bc+ca$.Rearranging the terms,we get: $a^{2}+b^{2}+c^{2}-ab-bc-ca=0$.Multiplying the entire equation by $2$,we get: $2a^{2}+2b^{2}+2c^{2}-2ab-2bc-2ca=0$.This can be rewritten as: $(a^{2}-2ab+b^{2})+(b^{2}-2bc+c^{2})+(c^{2}-2ca+a^{2})=0$.This simplifies to: $(a-b)^{2}+(b-c)^{2}+(c-a)^{2}=0$.Since the sum of squares is zero,each term must individually be zero:$a-b=0 \Rightarrow a=b$,$b-c=0 \Rightarrow b=c$,$c-a=0 \Rightarrow c=a$.Thus,$a=b=c$.Substituting these values into the expression $\frac{a+c}{b}$:$\frac{a+c}{b} = \frac{a+a}{a} = \frac{2a}{a} = 2$.

If $a^{2}+b^{2}+c^{2}=ab+bc+ac,$ then the value of $\frac{a+c}{b}$ is:

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