Factorise 27a3+b3+18c3_18abc using identity
Answers
Answered by
15
using a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)
27a³+b³+8c³-18abc = (3a)³+(b)³+(2c)³-3(3a)(b)(2c)
(3a+b+2c)(9a²+b²+4c²-3ab-2bc-6ac)
hope this helps
27a³+b³+8c³-18abc = (3a)³+(b)³+(2c)³-3(3a)(b)(2c)
(3a+b+2c)(9a²+b²+4c²-3ab-2bc-6ac)
hope this helps
Answered by
3
Answer:
27a³+b³+18c³+18abc
Step-by-step explanation:
(3a) ³+b³+(2c) ³-3*3a*b*2c
(3a+b+2c)(9a²+b²+2c²-3ab-2bc-6ca)
Similar questions