$4 a^{2} - (4 b^{2} + 4 b c + c^{2}$ )

$(2 a - 2 b - c) (2 a + 2 b + c)$

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$(2 a + 2 b - c) (2 a + 2 b + c)$

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$(2 a - 2 b - c) (2 a + 2 b - c)$

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$(2 a - 2 b - c) (2 a - 2 b + c)$

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Solution

The correct option is A
$(2 a - 2 b - c) (2 a + 2 b + c)$

Given that

$4 a^{2} - (4 b^{2} + 4 b c + c^{2}$ )
= $(2 a)^{2} - ((2 b + c)^{2}$
= $(2 a - (2 b + c)) (2 a + 2 b + c)$ ( using difference of squares )
= $(2 a - 2 b - c) (2 a + 2 b + c)$

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