How do you solve $x^{4} - 81$ $x^4-81$ ?

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How do you solve $x^{4} - 81$ $x^4-81$ ?

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Answer 1 · 2016-12-01T17:46:32

Resolve by factorization
$(x + 3) (x - 3) (x^{2} + 9)$ $(x+3)(x-3)(x^2+9)$

Explanation:

Resolve by factorization considering $a^{2} - b^{2} = (a + b) (a - b)$ $a^2 - b^2 = (a+b)(a-b)$
Detail is below
$x^{4} - 3^{4} = {(x^{2})}^{2} - {(3^{2})}^{2} = (x^{2} - 3^{2}) (x^{2} + 3^{2}) = (x + 3) (x - 3) (x^{2} + 9)$ $x^4-3^4 = (x^2)^2-(3^2)^2=(x^2-3^2)(x^2+3^2)=(x+3)(x-3)(x^2+9)$

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How do you solve $x^{4} - 81$ $x^4-81$ ?

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How do you solve $x^{4} - 81$ $x^4-81$ ?