Question

Factorize (256x^16 - 1)

Answer 1

256x^16-1
=> (16x^8)^2-1^2
=> (16x^8+1)(16x^8-1)
=> (16x^8+1)[(4x^4)^2-1^2)]
=> (16x^8+1)(4x^4+1)(4x^4-1)
=> (16x^8+1)(4x^4+1)[(2x^2)^2-1^2]
=> (16x^8+1)(4x^4+1)(2x^2+1)(2x^2-1)

Answer 2

Answer:

$256x^{16} - 1 = (16x^8+1)(4x^4+1)(2x^2+1)(2x^2-1)$

Step-by-step explanation:

Given : $256x^{16} - 1$

To Find : Factorize

Solution:

$256x^{16}-1$

$16x^8)^2-1^2$

$(16x^8+1)(16x^8-1)$

$(16x^8+1)[(4x^4)^2-1^2)]$

$(16x^8+1)(4x^4+1)(4x^4-1)$

$(16x^8+1)(4x^4+1)[(2x^2)^2-1^2]$

$(16x^8+1)(4x^4+1)(2x^2+1)(2x^2-1)$

Hence  $256x^{16} - 1 = (16x^8+1)(4x^4+1)(2x^2+1)(2x^2-1)$