Factorise :
64 - a3b3
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Given,
= (4)3 - (ab)3
= (4 - ab)[(4)2 + 4 × ab + (ab)2]
= (4 - ab)(16 + 4ab + a2b2).
Hence, 64 - a3b3 = (4 - ab)(16 + 4ab + a2b2).
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8a3 + 1 in the form of factors is :
(2a + 1)(4a2 - 2a + 1)
(2a - 1)(4a2 - 2a + 1)
(2a + 1)(4a2 + 2a + 1)
(2a - 1)(4a2 + 2a + 1)
a6 + 27b3
3x7y - 81x4y4