How do you factor $1 - 256 y^{8}$ $1 - 256y^8$ ?

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Answer 1 · 2015-07-15T07:51:29

Use the difference of squares identity three times to find:

$1 - 256 y^{8} = (1 - 2 y) (1 + 2 y) (1 + 4 y^{2}) (1 + 16 y^{4})$ $1-256y^8=(1-2y)(1+2y)(1+4y^2)(1+16y^4)$

$1 - 256 y^{8}$ $1-256y^8$

$= 1^{2} - {(16 y^{4})}^{2}$ $=1^2-(16y^4)^2$

$= (1 - 16 y^{4}) (1 + 16 y^{4})$ $=(1-16y^4)(1+16y^4)$

$= (1^{2} - {(4 y^{2})}^{2}) (1 + 16 y^{4})$ $=(1^2-(4y^2)^2)(1+16y^4)$

$= (1 - 4 y^{2}) (1 + 4 y^{2}) (1 + 16 y^{4})$ $=(1-4y^2)(1+4y^2)(1+16y^4)$

$= (1^{2} - {(2 y)}^{2}) (1 + 4 y^{2}) (1 + 16 y^{4})$ $=(1^2-(2y)^2)(1+4y^2)(1+16y^4)$

$= (1 - 2 y) (1 + 2 y) (1 + 4 y^{2}) (1 + 16 y^{4})$ $=(1-2y)(1+2y)(1+4y^2)(1+16y^4)$

...using the difference of squares identity:

$a^{2} - b^{2} = (a - b) (a + b)$ $a^2-b^2 = (a-b)(a+b)$

How do you factor 1 - 256y^81−256y81 - 256y^8?