If $a + b + c = 6$ and $a b + b c + c a = 11$ , find the value of $a^{2} + b^{2} + c^{2}$

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Solution

We have ${(a + b + c)}^{2} = a^{2} + b^{2} + c^{2} + 2 (a b + b c + c a)$
$\Rightarrow a^{2} + b^{2} + c^{2} = {(a + b + c)}^{2} - 2 (a b + b c + c a)$
$\Rightarrow a^{2} + b^{2} + c^{2} = {(6)}^{2} - 2 (11)$
$\Rightarrow a^{2} + b^{2} + c^{2} = 36 - 22 = 14$

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Similar questions

If a + b + c = 11 and $a^{2} + b^{2} + c^{2} = 81$ , find ; $a b + b c + c a$ .

\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}

a + b + c = 11

a b + b c + c a = 36

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

a + b + c = 0

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

If $a + b + c = 0$ then find value of $\frac{a^{2}}{b c} + \frac{b^{2}}{a c} + \frac{c^{2}}{a b}$

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