If a+b+c=6, a^{2}+b^{2}+c^{2}=14 and a^{3}+b^ | vedclass

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(B) We know the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$.Substituting the given values: $6^{2} = 14 + 2(ab+bc+ca)$.$36 = 14 + 2(ab+bc+ca

Vedclass · Accepted Answer

$6$

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(B) We know the identity $(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$.Substituting the given values: $6^{2} = 14 + 2(ab+bc+ca)$.$36 = 14 + 2(ab+bc+ca) \Rightarrow 2(ab+bc+ca) = 22 \Rightarrow ab+bc+ca = 11$.Next,we use the identity $a^{3}+b^{3}+c^{3}-3abc = (a+b+c)(a^{2}+b^{2}+c^{2}-(ab+bc+ca))$.Substituting the known values: $36 - 3abc = 6(14 - 11)$.$36 - 3abc = 6(3) = 18$.$3abc = 36 - 18 = 18$.$abc = 6$.

If $a+b+c=6, a^{2}+b^{2}+c^{2}=14$ and $a^{3}+b^{3}+c^{3}=36,$ then the value of $abc$ is

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