Without using long division or synthetic division show how x^2-4 is a factor of x^6-64.?

a³ - b³ = (a - b)(a² + ab + b²)

Put a = x² and b = 4

x⁶ - 64
= (x²)³ - (4)³
= a³ - b³
= (a - b)(a² + ab + b²)
= (x² - 4)[(x²)² + (x²)(4) + (4)²]
= (x² - 4)(x⁴ + 4x² + 16)

Hence, x² - 4 is a factor of x⁶ - 64.

(a-b)³ = a³ - 3a²b + 3ab² - b³ = a³ - b³ - 3ab(a-b)
so a³ - b³ = (a-b)³ + 3ab(a-b) = ((a-b)² + 3ab)(a-b), so a-b is a factor of a³-b³.

Letting a = x^2 (so a³ = x^6) and b=4 (so b³ = 64) we have x^2-4 is a factor of x^6-64.

x^6-64 = (x^2)^3-4^3
= (x^2-4)(x^4+4x^2+16)

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

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

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

2 and -2 are zeroes of both expressions. So x^2-4 = (x-2)(x+2) must be a factor of x^6 - 64.

Another option is to note that the roots of x^2 - 4 are 2 and -2, and those are easily seen to be roots of x^6 - 64.

x⁶ – 64
use difference of squares
(x³ + 8)(x³ – 8)
factor each
(x + 2)(x² – 2x + 4)(x – 2)(x² + 2x + 4)
use difference of squares again, reverse
(x² – 4)(x² – 2x + 4)(x² + 2x + 4)


rules:
a² – b² = (a + b)(a – b)
a³ + b³ = (a + b)(a² – ab + b²)
a³ – b³ = (a – b)(a² + ab + b²)

Put the value x^2=4 and see that x^6-64 becomes 0. If it is then x^2-4=0 is one of the solution.

x^6 - 64 = (x^3 - 8)(x^3 + 8)
= (x - 2)(x^2 + 2x + 4)(x + 2)(x^2 - 2x + 4)
= (x^2 - 4)(x^2 + 2x + 4)(x^2 - 2x + 4)

By the difference of two squares,
x² - 4 = (x - 2)(x + 2)
ie x² - 4 has roots of +2 and -2

By inspection +2 and -2 are roots of x⁶ - 64,
thus (x - 2) and (x + 2) are each factors of x⁶ - 64,
so (x - 2)(x + 2) is a factor too (= x² - 4 )

x^6 - 64
= (x - 2) (x + 2) (x^2 - 2 x + 4) (x^2 + 2 x + 4)
= (x^2 - 4)(x^2 - 2 x + 4) (x^2 + 2 x + 4)