Question

if a+b+c=9and ab+bc+ca=26,find a²+b²+c²
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Answer 1

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: \: a + b + c = 9$

$\: \: \: \: \: \: \: \: \: \: \: ab + bc + ca = 26$

$(a + b + c) {}^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2ab + 2bc + 2ca$

$a {}^{2} + b {}^{2} + c {}^{2} = (a + b + c) {}^{2} - 2ab - 2bc - 2ca$

$\: : ⟹ \: {9}^{2} - 2(ab + bc + ca)$

$\: \: \: \: \: : ⟹ \: \: \: \: 81 - 2(26)$

$\: \: \: \: \: \: \: \: \: \: : ⟹ \: 81 - 52$

$\: \: \: \: \: \: \: \: \: \: \: \: \: \: : ⟹ \: \green{ \boxed{29}}$

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Answer 2

Answer:

We know (a+b+c)

2

=a

2

+b

2

+c

2

+2ab+2bc+2ca

⇒(a+b+c)

2

=a

2

+b

2

+c

2

+2(ab+bc+ca)

Given, a+b+c=9 and ab+bc+ca=26

Putting these values, we get

(9)

2

=a

2

+b

2

+c

2

+2(26)

⇒81=a

2

+b

2

+c

2

+52

⇒a

2

+b

2

+c

2

=81−52=29