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Algebraic Identity 4

Simplify: a+b...

Question

Simplify: $(a + b + c) (a + b - c)$

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

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

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

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

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Solution

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

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

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)

a^{2} + b^{2} + c^{2} - a b - b c - c a = \frac{1}{2} [{(b - c)}^{2} + {(c - a)}^{2} + {(a - b)}^{2}]

In a triangle ABC, $2 c a s i n \frac{A - B + C}{2}$ is equal to

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