How do you factor $x^{4 n} - y^{4 n}$ $x^(4n) - y^(4n)$ ?

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How do you factor $x^{4 n} - y^{4 n}$ $x^(4n) - y^(4n)$ ?

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Answer 1 · 2016-05-12T00:12:41

$(x^{2 n} + y^{2 n}) (x^{n} + y^{n}) (x^{n} - y^{n})$ $(x^(2n)+y^(2n))(x^n+y^n)(x^n-y^n)$

Explanation:

Note that both $x^{4 n}$ $x^(4n)$ and $y^{4 n}$ $y^(4n)$ are squared terms:

$x^{4 n} = {(x^{2 n})}^{2}$ $x^(4n)=(x^(2n))^2$
$y^{4 n} = {(y^{2 n})}^{2}$ $y^(4n)=(y^(2n))^2$

With this knowledge, we will factor this expression as a difference of squares:

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

Thus, we see that

$x^{4 n} - y^{4 n}$ $x^(4n)-y^(4n)$

$= {(x^{2 n})}^{2} - {(y^{2 n})}^{2}$ $=(color(red)(x^(2n)))^2-(color(blue)(y^(2n)))^2$

$= (x^{2 n} + y^{2 n}) (x^{2 n} - y^{2 n})$ $=(color(red)(x^(2n))+color(blue)(y^(2n)))(color(red)(x^(2n))-color(blue)(y^(2n)))$

Notice that we can factor $(x^{2 n} - y^{2 n})$ $(x^(2n)-y^(2n))$ in the same way:

$x^{4 n} - y^{4 n} = (x^{2 n} + y^{2 n}) (x^{2 n} - y^{2 n})$ $x^(4n)-y^(4n)=(x^(2n)+y^(2n))(x^(2n)-y^(2n))$

$= (x^{2 n} + y^{2 n}) ({(x^{n})}^{2} - {(y^{n})}^{2})$ $=(x^(2n)+y^(2n))((x^n)^2-(y^n)^2)$

$= (x^{2 n} + y^{2 n}) (x^{n} + y^{n}) (x^{n} - y^{n})$ $=(x^(2n)+y^(2n))(x^n+y^n)(x^n-y^n)$

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How do you factor $x^{4 n} - y^{4 n}$ $x^(4n) - y^(4n)$ ?

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