Divide:
$[(a^{2} + 2 a b + b^{2}) - (a^{2} + 2 a c + c^{2})]$ by $2 a + b + c$

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Solution

Given $[(a^{2} + 2 a b + b^{2}) - (a^{2} + 2 a c + c^{2})] \div (2 a + b + c)$
$= [(a + b)^{2} - (a + c)^{2}] \div (2 a + b + c)$ [Since, $a^{2} + 2 a b + b^{2} = (a + b)^{2}, a^{2} - b^{2} = (a - b) (a + b)$ ]
$= [(a + b + a + c) (a + b - a - c)] \div (2 a + b + c)$
$= (2 a + b + c) (b - c) \div (2 a + b + c)$
$∴ [(a^{2} + 2 a b + b^{2}) - (a^{2} + 2 a c + c^{2})] \div (2 a + b + c) = (b - c)$

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Similar questions

If a + b + c = 2s, then prove the following identities

(a) s² + (s − a)² + (s − b)² + (s − c)² = a² + b² + c²

(b) a² + b² − c² + 2ab = 4s (s − c)

(d) a² − b² − c² + 2ab = 4(s − b) (s − c)

(e) (2bc + a² − b² − c²) (2bc − a² + b² + c²) = 16s (s − a) (s − b) (s − c)

(f)

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

a, b,

c

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

Δ A B C

c o s A = \frac{b^{2} + c^{2} - a^{2}}{2 b c}, c o s B = \frac{c^{2} + a^{2} - b^{2}}{2 a c}, c o s C = \frac{a^{2} + b^{2} - c^{2}}{2 a b}

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