Question

Factorise: (i) 1 − 64x^3​

Answer 1

1 − 64x^3

= (1)^3 - (4x)^3

= (1 - 4x) (1 + 4x + 16x^2)

Answer 2

Answer:

Factor the following:

• $\mathbb{ \red{-64 x^3 + 1}}$

Factor -1 out of -64 x^3 + 1:

• -(64 x^3 - 1)

• 64 x^3 - 1 = (4 x)^3 - 1^3:
• -((4 x)^3 - 1^3)

• Factor the difference of two cubes. (4 x)^3 - 1^3 = (4 x - 1) ((4 x)^2 + 4 x + 1^2):

• -((4 x - 1) ((4 x)^2 + 4 x + 1^2))

1^2 = 1:

• -(4 x - 1) ((4 x)^2 + 4 x + 1)

Multiply each exponent in 4 x by 2:

• -(4 x - 1) (16 x^2 + 4 x + 1)

4^2 = 16:

Answer: |

• | -(4 x - 1) (16 x^2 + 4 x + 1)

Step-by-step explanation:

$\mathbb{ \huge{ \color{lime}-(4 x - 1) (16 x^2 + 4 x + 1)}}$