Question

if a+b+c=9 and ab+bc+ca=40,find a square +b sqaure+c square

Answer 1

Hi there !!

Given,

a + b + c = 9

ab + bc + ca = 40

To find : a² + b² + c²

The following expression matches to the algebraic identity
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac or
(a + b + c)² = a² + b² + c² + 2( ab + bc + ac)

Thus,

(a + b + c)² = a² + b² + c² + 2(ab + bc + ac)

substituting the values,
we have,

9² = a² + b² + c² + 2( 40)

81 = a² + b² + c² + 80

Transposing 80 to LHS,
we have,

81 - 80 = a² + b² + c²

Thus,

a² + b² + c² = 1

Answer 2

$(a + b + c)^{2} = a^{2} + {b}^{2} + c^{2} + 2ab + 2bc+ 2ac \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = {a}^{2} + {b}^{2} + \:{c}^{2} + 2(ab + bc + ca) \\ given \: ab + bc + ca = 40 \: and \: a + b + c = 9 \\ ({9})^{2} = {a}^{2} + {b}^{2} + {c}^{2} + 2(40) \\ {a}^{2} + {b}^{2} + {c}^{2} = 81 - 80 \\ {a}^{2} + {b}^{2} + {c}^{2} = 1$
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