Factorise:
$a^{2} - 2 a b + b^{2} - c^{2}$

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Solution

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

$= {(a - b)}^{2} - c^{2} (∵ {(x - y)}^{2} = x^{2} - 2 x y + y^{2})$

$= (a - b + c) (a - b - c) (∵ x^{2} - y^{2} = (x - y) (x + y))$

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Similar questions

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

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

Factorise:

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

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

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