Question

factorise (ii) 14(a – 3b)3 – 21p(a – 3b)​

Answer 1

Answer:

14(a−3b)

3

−21p(a−3b)

Take out common in all terms,

Then,7(a−3b)[2(a−3b)

2

−3p]

Therefore, HCF of 14(a−3b)

3

and 21p(a−3b) is 7(a−3b).

Answer 2

Answer:

(a - 3b) (42 - 21p)

Step-by-step explanation:

14(a – 3b)3 – 21p(a – 3b)​

= 14 × 3(a - 3b) - 21p(a - 3b)

= 42(a - 3b) - 21p(a - 3b)

= (a - 3b) (42 - 21p)

Hope that helps you...☺