If $(a + b + c = 25)$ and $(a b + b c + c a = 59)$

Find the value of

$(a^{2} + b^{2} + c^{2})$

118

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500

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527

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507

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Solution

The correct option is D
507

According to the question,

$a + b + c = 25$

Squaring both the sides, we get

$(a + b + c)^{2} = 25^{2})$

$a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a = 625$

$a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a) = 625$

$a^{2} + b^{2} + c^{2} + 2 \times 59 = 625 [G i v e n, a b + b c + c a = 59]$

$a^{2} + b^{2} + c^{2} + 118 = 625$

$(a^{2} + b^{2} + c^{2} + 118) - 118 = 625 - 118 [s u b t r a c t i n g 118 f r o m b o t h t h e s i d e s]$

$∴ a^{2} + b^{2} + c^{2} = 507$

Suggest Corrections

Similar questions

\frac{a^{2} + b^{2} + a b}{b^{2} + c^{2} + b c} + \frac{c^{2} + c a + a^{2}}{b^{2} + c^{2} + b c}

\frac{a^{2}}{b c} + \frac{b^{2}}{c a} + \frac{c^{2}}{a b}

a + b + c = 0

\frac{(b + c)^{2}}{b c} + \frac{(c + a)^{2}}{c a} + \frac{(a + b)^{2}}{a b}

a^{2} + b^{2} + c^{2} = 31

a b + b c + c a = 25

a + b + c

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