contestada


1.On factorising (3a^2bc + 9ab²c + 21abc^2), we get :

(a) 3abc (3a + 3b + 7c)
(b) 3abc (a + b + c)
(c) abc (a + 3b + 7c)
(d) 3abc (a + 3b + 7c)

Respuesta :

konnievas
actually it’s none of these.

24a^2bc+9ab2c
48abc+9ab2c

answer: 3abc ^ (16+3b)
1774928

Answer:

Fast

Step-by-step explanation:

You told me "Please answer fast." ;-;