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Question
Factorise : 64 - a3b3
Sum
Solution
64 - a3b3
= (4)3 - (ab)3
= ( 4 - ab )[(4)2 + 4 x ab + (ab)2 ] [ ∵ a3 - b3 = ( a - b )( a2 + ab + b2 )]
= ( 4 - ab )( 16 + 4ab + a2b2 )
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Method of Factorisation : the Sum Or Difference of Two Cubes
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