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Question
Factorise: 4a2 - (4b2 + 4bc + c2)
Sum
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Solution
4a2 - (4b2 + 4bc + c2)
= (2a)2 - [(2b)2 + (c)2 + 2 × 2b × c]
= (2a)2 - (2b + c)2
= [2a - (2b + c)][2a + (2b + c)] ...[∵ a2 - b2 = (a + b)(a - b)]
= [2a - 2b - c][2a + 2b + c]
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Method of Factorisation : Difference of Two Squares
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