Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

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Solution

Consider $(a - b)^{2} = (a - b) (a - b)$
$= a (a - b) - b (a - b)$
$= (a \times a) - (a \times b) - (b \times a) + (b \times b$ )
$= a^{2} - a b - a b + b^{2}$
$= a^{2} - 2 a b + b^{2}$ [because $a . b = b . a$ ]
and also $(a + b)^{2} = (a + b) (a + b)$
$= (a \times a) + (a \times b) + (b \times a) + (b \times b)$
$= a^{2} + 2 a b + b^{2}$
Hence, the only correct option is (b).

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

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(a - b)^{2} = a^{2} - b^{2}

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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)

Find:
$(a + b + c) \times (a + b - c) =$

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

(a^{2} + b^{2} +2ab) - (a^{2} + b^{2} - 2ab)

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