we have
Part (1):-
64x3−729y3
(4x)3−(9y)3
Using (x3−y3)=(x−y)(x2+y2+xy)
(4x)3−(9y)3=(4x−9y)[(4x)2+(9y)2+4x×9y]
=(4x−9y)(16x2+81y2+36xy)
Evaluate : ( 8a³ - 27b³ ) / ( 64x³ + y³ ) , where ( 64x³ + y³ ) ≠ 0
Evaluate (8a3−27b3)(64x3+y3), where (64x3+y3)≠0.
Factorise: x−64x3