Question

factorise the following (1) 1-64a³-12a +48a²​

Answer 1

Answer:

1-64a^3-12a+48a^2

=(1)^3-(4a)^3-12a(1-4a)

=(1-4a){(1)^2+1*4a+(4a)^3}-12a(1-4a)

=(1-4a)(1+4a+16a^2)-12a(1-4a)

=(1-4a)(1+4a+16a^2-12a)

=(1-4a)(1-8a+16a^2)

=(1-4a){(1)^2-2.1.4a+(4a)^2}

=(1-4a)(1-4a)^2

=(1-4a)^3