Divide-a2 - 2ab - b2- minus- -a2 - 2ac - c2- by 2a - b - c

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

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Division of an Expression by a Monomial

Divide: (a2 +...

Question

Divide:
(a² + 2ab + b²) − (a² + 2ac + c²) by 2a + b + c

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Solution

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

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

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