Question

10. Factorise by using the above results
(identities)
1) 27a3+ 64b3​

Answer 1

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=

(3a)3−43b3=(3a)3−(4b)3

Now we can apply the rule:

27a3−64b3=(3a)3−(4b)3=

=(3a−4b)((3a)2+12ab+(4b)2)

=(3a−4b)(9a2+12ab+16b

Answer 2

Answer:

27a³+ 64b³

= (3a)³+(4b)³

= (3a+8b){(3a)²-3a.4b+(4b)²}

= (3a+8b)( 9a²-12ab+16b²)

hope it works out for you....