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10. Factorise by using the above results(identities)1) 27a3+ 64b3 |
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Answer» Hope this helps you. Step-by-step explanation: Remembering that: a3−b3=(a−b)(a2+ab+b2) we can try to write 27a3−64b3 like a difference of cubes 27a3−64b3=33a3−26b3=33a3−(22)3b3= Now we can apply the rule: 27a3−64b3=(3a)3−(4b)3= =(3a−4b)((3a)2+12ab+(4b)2) =(3a−4b)(9a2+12ab+16b |
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