Prove that
$(a - b)^{2} = (a + b)^{2} - 4 a b$

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Solution

$L H S = (a - b)^{2}$
$= a^{2} + b^{2} - 2 a b = a^{2} + b^{2} + 2 a b - 4 a b = (a + b)^{2} - 4 a b = R H S$

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Similar questions

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

(a + b)^{2} - (a - b)^{2} = 4 a b

(a + b)^{2} = 4 a b

a = b

{(a + b)}^{2} - {(a - b)}^{2} = 4 a b

(2 a + b)^{2} = 4 a^{2} + b^{2} + 4 a b

{(a - b)}^{2}

4 a b

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