How do you factor $x^16-81$ ?

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Answer 1 · 2017-07-14T13:55:13

$(x - x_0)(x - x_1)(x - x_2) ... (x - x_15)$

When we solve $z^16 = 81$ , we find 16 roots.

$x_n = 81^(1/16) (cos frac{2 pi * n}{16} + i sin frac{2 pi * n}{16})$

$x_n = 3^(1/4) (cos frac{pi * n}{8} + i sin frac{pi * n}{8})$

Answer 2 · 2017-07-14T13:55:33

As the difference of two squares $(x^8 -9) xx (x^8 +9)$

$x^16$ can be consider the square of $x^8 xx x^8$

81 can be consider the square of $9xx 9$

when the binomials of two squares are multiplied as alternatively positive and negative terms the middle term falls out.

$( x^8 -9) xx (x^8 + 9 ) = x^16 - 9x^8 + 9x^8 - 81$

$-9x^8 + 9x^8 = 0$ so

$(x^8 -9) xx (x^8 +9) = x^16 -81$

Answer 3 · 2017-07-14T15:00:40

$(x^8+9)((x^4+3)(x^4-3)$

$x^16-81$

using difference of squares

$=(x^8+9)(x^8-9)$

using DoS on the second bracket

$(x^8+9)((x^4+3)(x^4-3)$

How do you factor x^16-81x^16-81?