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20 March, 16:32

Factor completely 81x8 - 1. (9x4 - 1) (9x4 + 1) (3x2 - 1) (3x2 + 1) (9x4 - 1) (3x2 - 1) (3x2 + 1) (9x4 + 1) (3x2 - 1) (3x2 + 1) (3x2 + 1) (3x2 + 1)

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  1. 20 March, 16:43
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    (3x^2 - 1) (3x^2 + 1) (9x^4 + 1).

    Step-by-step explanation:

    Using the identity for the difference of 2 squares;

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

    we put a^2 = 81x^8 and b^2 = 1 giving

    a = 9x^4 and b = 1, so:

    81x^8 - 1 = (9x^4 - 1) (9x^4 + 1)

    Applying the difference of 2 squares to 9x^4 - 1:

    = (3x^2 - 1) (3x^2 + 1) (9x^4 + 1).
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