Question

use identity to find (x^256-1)​

Answer 1

ANSWER:

To Find:

• x^(256) - 1

Solution:

We will use the identity,

$\hookrightarrow a^2-b^2=(a+b)(a-b)$

We are given that,

$\implies x^{256}-1$

We can write it as,

$\implies x^{128\times2}-1$

$\implies (x^{128})^2-1^2$

Using the above mentioned formula,

$\implies (x^{128})^2-1^2$

$\implies \left(x^{128}-1\right)\left(x^{128}+1\right)$

$\implies \left(x^{64\times2}-1\right)\left(x^{128}+1\right)$

$\implies \bigg(\left(x^{64}\right)^2-1^2\bigg) \bigg(x^{128}+1\bigg)$

Using the mentioned formula,

$\implies \bigg(\left(x^{64}\right)^2-1^2\bigg) \bigg(x^{128}+1\bigg)$

$\implies \bigg(x^{64}-1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(x^{32\times2}-1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(\left(x^{32}\right)^2-1^2\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

Using the mentioned formula,

$\implies \bigg(\left(x^{32}\right)^2-1^2\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(x^{32}-1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(x^{16\times2}-1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(\left(x^{16}\right)^2-1^2\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

Using the mentioned formula,

$\implies \bigg(\left(x^{16}\right)^2-1^2\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(x^{16}-1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(x^{8\times2}-1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(\left(x^8\right)^2-1^2\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

Using the mentioned formula,

$\implies \bigg(\left(x^8\right)^2-1^2\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(x^8-1\bigg)\bigg(x^8+1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(x^{4\times2}-1\bigg)\bigg(x^8+1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(\left(x^4\right)^2-1^2\bigg)\bigg(x^8+1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

Using the mentioned formula,

$\implies \bigg(\left(x^4\right)^2-1^2\bigg)\bigg(x^8+1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(x^4-1\bigg)\bigg(x^4+1\bigg)\bigg(x^8+1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(x^{2\times2}-1\bigg)\bigg(x^4+1\bigg)\bigg(x^8+1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

$\implies \bigg(\left(x^2\right)^2-1^2\bigg)\bigg(x^4+1\bigg)\bigg(x^8+1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

Using the mentioned formula,

$\implies \bigg(\left(x^2\right)^2-1^2\bigg)\bigg(x^4+1\bigg)\bigg(x^8+1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

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

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

Using the mentioned formula,

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

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

Therefore,

$\implies\bf x^{256}-1=\bigg(x-1\bigg)\bigg(x+1\bigg)\bigg(x^2+1\bigg)\bigg(x^4+1\bigg)\bigg(x^8+1\bigg)\bigg(x^{16}+1\bigg)\bigg(x^{32}+1\bigg)\bigg(x^{64}+1\bigg)\bigg(x^{128}+1\bigg)$

Answer 2

$reffered \: to \: the \: attachment$