Question

factorize using a3-b3 and a3+b3 in 8p3-27q2​

Answer 1

Step-by-step explanation:

Notice the terms are both perfect cubes

x3 = (x)3

27y9 = (3y3)3

 x3 + 27y9 = (x)3 + (3y3)3

a3 + b3

and we have a sum

= (x + 3y3)((x)2 - (x)(3y3) + (3y3)2)

 sum of cubes

= (a + b)(a2 - ab + b2)

factors as

= (x + 3y3)(x2 – 3xy3 + 9y6)

Slide 59

Answer 2

Answer:

that's not possible because there is square in 27q

if there would be cube

Step-by-step explanation:

8p³ — 27q³

(2p — 3q)³

hence factories

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