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Question
Factorise : (a2 + b2 - 4c2)2 - 4a2b2
Sum
Solution
(a2 + b2 - 4c2)2 - 4a2b2
= ( a2 + b2 - 4c2 )2 - ( 2ab )2
= ( a2 + b2 - 4c2 - 2ab )( a2 + b2 - 4c2 + 2ab ) [ ∵ a2 - b2 = ( a + b )( a - b )]
= ( a2 + b2 - 2ab - 4c2 )( a2 + b2 + 2ab - 4c2 )
= [ ( a - b )2 - ( 2c )2 ][ ( a + b )2 - ( 2c )2]
= ( a - b + 2c )( a - b - 2c )( a + b + 2c )( a + b - 2c )
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Method of Factorisation : Difference of Two Squares
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