Question

use the identity to find the value(x^256-1). Its urgent ​

Answer 1

You may find the answer below:-

$x^{256} -1$

= $(x^{128})^2-1$

applying identity, $a^{2}-b^2=(a+b)(a-b)$ and $1^2=1$ .

= $(x^{128} + 1) ( x^{128} - 1)$

= $(x^{128} + 1) ((x^{64})^2-1)$

= $(x^{128} + 1) ( x^{64} - 1) ( x^{64} + 1)$

= $(x^{128} + 1) ( x^{64} + 1) ((x^{32})^2-1)$

= $(x^{128} + 1) ( x^{64} + 1) ( x^{32} + 1) (x^{32}-1)$

if you go on at last you get,

= $(x^{128} + 1) ( x^{64} + 1) ( x^{32} + 1) (x^{16}+1)(x^{8}+1)(x^{4}+1)(x^{2}+1)(x+1)(x-1)$

Pls mark brainliest as it took time with math inputing :D.