If a+b+c=1 and ab+bc+ca=frac{1}{3},then a:b:c | vedclass

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(C) Given equations are $a+b+c=1$ and $ab+bc+ca=\frac{1}{3}$.We know the identity $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$.Substituting the given values:

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$1:1:1$

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(C) Given equations are $a+b+c=1$ and $ab+bc+ca=\frac{1}{3}$.We know the identity $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca)$.Substituting the given values: $(1)^2 = a^2+b^2+c^2+2(\frac{1}{3})$.$1 = a^2+b^2+c^2+\frac{2}{3} \Rightarrow a^2+b^2+c^2 = 1-\frac{2}{3} = \frac{1}{3}$.Now,consider the expression $(a-b)^2+(b-c)^2+(c-a)^2 = 2(a^2+b^2+c^2) - 2(ab+bc+ca)$.Substituting the values: $2(\frac{1}{3}) - 2(\frac{1}{3}) = 0$.Since the sum of squares is zero,each term must be zero: $a-b=0, b-c=0, c-a=0$.Thus,$a=b=c$.Since $a+b+c=1$,we have $3a=1$,so $a=b=c=\frac{1}{3}$.Therefore,the ratio $a:b:c = \frac{1}{3}:\frac{1}{3}:\frac{1}{3} = 1:1:1$.

If $a+b+c=1$ and $ab+bc+ca=\frac{1}{3}$,then $a:b:c$ is

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