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प्रश्न
Factorise : 4a2 - (4b2 + 4bc + c2)
बेरीज
उत्तर
4a2 - (4b2 + 4bc + c2)
= ( 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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