Can anyone factor 625x^4-1?

(25x^2 + 1)(25x^2 - 1) =

(25x^2 + 1)(5x + 1)(5x - 1). This as far as things go if you just want to factor over the reals. If you want to factor over all numbers (including imaginary numbers), you can go further.

(25x^2 + 1)(5x + 1)(5x - 1) =

(5x + i)(5x - i)(5x + 1)(5x - 1).

(25 x ² + 1)(25 x ² - 1)
(25 x ² + 1)(5x - 1)(5x + 1)

a^2 - b^2 = (a + b)(a - b)

625x^4 - 1
= (25x^2)^2 - 1^2
= (25x^2 + 1)(25x^2 - 1)
= (25x^2 + 1)[(5x)^2 - 1^2]
= (25x^2 + 1)(5x + 1)(5x - 1)

625x^4-1 = (25x^2 - 1) (25x^2+1) = (5x+1)(5x-1)(5x-i)(5x+i)

625 = 25 x 25
factorise using the difference of two squares, twice :
625x^4-1 = (25x^2)^2 - 1^2
= (25x^2 - 1)(25x^2 + 1)
= (5x - 1)(5x+1)(25x^2 + 1)

(25x^2 + 1)(25x^2 - 1)
(25x^2 + 1)(5x + 1)(5x - 1)

Easy......(25x^2 + 1)(5x + 1)(5x - 1)

(5x-1)(5x+1)(25x²+1)