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Factor completely 16x^8 - 1.

(4x^4 - 1) (4x^4 + 1)

(2x^2 - 1) (2x^2 + 1) (4x^4 + 1)

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

(2x^2 - 1) (2x^2 + 1) (4x^4 - 1)

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  1. 26 May, 06:54
    0
    16x^8 - 1 = (4x^4 + 1) (2x^2 + 1) (2x^2 - 1)

    Step-by-step explanation:

    The difference of squares:

    ((ax) ^2n - b) = (ax^n + b) (ax^n - b)

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

    (4x^4 - 1) = (2x^2 + 1) (2x^2 - 1)

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