If a+b+c=9 and ab+bc+ca=26,find a^{2}+b^{2}+c | vedclass

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(B) We know the algebraic identity:$(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$Given that $a+b+c = 9$ and $ab+bc+ca = 26$.Substituting these values i

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$29$

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(B) We know the algebraic identity:$(a+b+c)^{2} = a^{2}+b^{2}+c^{2}+2(ab+bc+ca)$Given that $a+b+c = 9$ and $ab+bc+ca = 26$.Substituting these values into the identity:$(9)^{2} = (a^{2}+b^{2}+c^{2}) + 2(26)$$81 = (a^{2}+b^{2}+c^{2}) + 52$Subtracting $52$ from both sides:$a^{2}+b^{2}+c^{2} = 81 - 52$$a^{2}+b^{2}+c^{2} = 29$

If $a+b+c=9$ and $ab+bc+ca=26$,find $a^{2}+b^{2}+c^{2}$.

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