Question

Resolve a³ - b³ + 1 + 3ab into factors. ​

Answer 1

Answer:

Step-by-step explanation:

(x3−y3)=(x−y)(x2+xy+y2)

(a+b)3−1=(a+b)3−(1)3=(a+b−1)((a+b)2+(a+b)+1)

so we have

(a+b)3−1−3ab(a+b−1)=(a+b−1)((a+b)2+(a+b)+1)−3ab(a+b−1)

or

(a+b-1)\left((a+b)^2+a+b+1-3ab\right) = (a+b-1)\left(a^2+b^2-ab+a+b+1)

so

a^3+b^3-1+3ab=(a+b-1)\left(a^2+b^2-ab+a+b+1)

 

Answer 2

$<b>$We have;

a³ - b³ + 1 + 3ab

= a³ + (-b) ³ + 1³ -3(a) (-b) (1)

= (a-b+1) (a²+b²+1+ab-a+b) = (a-b+1)(a²+b²+ab-a+b+1)

Hope it helped!!