Factor p^12-125 - Math help

$p^{12} - 125$

Rewrite $p^{12}$ as ${(p^{4})}^{3}$ .

${(p^{4})}^{3} - 125$

Rewrite $125$ as $5^{3}$ .

${(p^{4})}^{3} - 5^{3}$

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = p^{4}$ and $b = 5$ .

$(p^{4} - 5) ({(p^{4})}^{2} + p^{4} \cdot 5 + 5^{2})$

Simplify.

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Multiply the exponents in ${(p^{4})}^{2}$ .

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Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$(p^{4} - 5) (p^{4 \cdot 2} + p^{4} \cdot 5 + 5^{2})$

Multiply $4$ by $2$ .

$(p^{4} - 5) (p^{8} + p^{4} \cdot 5 + 5^{2})$

Move $5$ to the left of $p^{4}$ .

$(p^{4} - 5) (p^{8} + 5 p^{4} + 5^{2})$

Raise $5$ to the power of $2$ .

$(p^{4} - 5) (p^{8} + 5 p^{4} + 25)$

Factor p^12-125

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