Factor x^4-64=0

$x^{4} - 64 = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

${(x^{2})}^{2} - 64 = 0$

Rewrite $64$ as $8^{2}$ .

${(x^{2})}^{2} - 8^{2} = 0$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 8$ .

$(x^{2} + 8) (x^{2} - 8) = 0$

Factor x^4-64=0

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