Factor this equation: x^8 - 256?

= x⁸ - 256
= (x⁴ + 16)(x⁴ - 16)
= (x⁴ + 16)(x² + 4)(x² - 4)
= (x⁴ + 16)(x² + 4)(x + 2)(x - 2)

Answer: (x⁴ + 16)(x² + 4)(x + 2)(x - 2)

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

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

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

x^8 - 256
= x^8-2^8
= (x^4+2^4)(x^2+2^2)(x+2)(x-2)

x^8 - 256
=(x^4+16)(x^4-16)
=(x^4+16)(x^2+4)(x^2-4)
=(x^2+16)(x^2+4)(x+2)(x-2) answer//

a^2 - b^2 ≡ (a + b)(a - b)

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

x^8-256

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

So simple algebra factoring :)

Just use this rule of factoring difference of 2 squares (a^2 - b^2) = (a-b)(a+b)

we can write them as difference of two squares :
(x^4)^2 - 16 ^2
(x^4 - 16 ) (x^4 +16)

Now, you ask yourself, can we factorize more??
You see the first term is also a difference between two squars, but the second is a sum of two squares, and sum of two squares cannot be factorized...

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

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

Another difference between two squares,,,

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

Hope this helps =)