factorise the 1-64a^3-12a+48a^2 with the suitable identity
Answers
Answered by
1
Answer:
(1-4a)³
Step-by-step explanation:
-64a³-12a+48a²
=(1)³-(4a)³-3(1)²(4a)+3(1)(4a)²
This is in the form of a³-b³-3a²b+3ab².
So,a³-b³-3a²b+3ab² = (a-b)³
Here,
a=1,b=4a
So,
(1)³-(4a)³-3(1)²(4a)+3(1)(4a)² = (1-4a)³
So,answer is (1-4a)³
Similar questions