Question

Factorise: 1-64a³-12a+48a²
Plz guys I want the answer urgently

Answer 1

→1-64a³-12a+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

Plz mark as brainliest

Answer 2

=(1³)_(4a)³_3(1)²(4a)+3(1)(4a)²

( a³_b³_3a²b+3ab²=(a_b)³

a=1, b=4a

so,

(1)³_(4a)³_3(1)²(4a)+3(1)(4a)²

=(1_4a)²