How do you factor $13 {(x^{6} + 1)}^{4}$ $13(x^6+ 1)^4$ ?

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How do you factor $13 {(x^{6} + 1)}^{4}$ $13(x^6+ 1)^4$ ?

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Answer 1 · 2018-01-24T16:59:00

$13 {(x^{2} + 1)}^{4} {(x^{4} - x^{2} + 1)}^{4}$ $13 (x^2+1)^4 (x^4 - x^2 + 1)^4$

Explanation:

$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ $a^3 + b^3 = (a+b)(a^2-ab+b^2)$
$\Rightarrow x^{6} + 1 = (x^{2} + 1) (x^{4} - x^{2} + 1)$ $=> x^6 + 1 = (x^2 + 1)(x^4 - x^2 + 1)$
$\Rightarrow 13 {(x^{6} + 1)}^{4} = 13 {(x^{2} + 1)}^{4} {(x^{4} - x^{2} + 1)}^{4}$ $=> 13(x^6+1)^4 = 13 (x^2+1)^4 (x^4 - x^2 + 1)^4$

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How do you factor $13 {(x^{6} + 1)}^{4}$ $13(x^6+ 1)^4$ ?

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How do you factor $13 {(x^{6} + 1)}^{4}$ $13(x^6+ 1)^4$ ?