Question

Factorise : x^12 – 4096​

Answer 1

Answer:

(x^6+64)(x^6-64)

(x^6+64)((x^3+8)(x^3-8))=0

hence, the awnser is (x^6+64)((x^3+8)(x^3-8))

Answer 2

Given:

$\bold{x^{12} -4096}$

To find:

Factorise.

Solution:

$\Rightarrow x^{12} - 4096\\ \Rightarrow (x^6)^2 -(2^6)^2\\ \Rightarrow (x^6)^2 -(64)^2\\\Rightarrow (x^6 -64)(x^6+64)\\ \Rightarrow (x^3 -8) (x^3 +8)(x^6+64)$

To solve  the above equation it will give:

$\bold{\boxed{\Rightarrow (x^2+4)(x^4-4x^2+16)(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)}}$