Factor x^12-y^9 - Math solver

$x^{12} - y^{9}$

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

${(x^{4})}^{3} - y^{9}$

Rewrite $y^{9}$ as ${(y^{3})}^{3}$ .

${(x^{4})}^{3} - {(y^{3})}^{3}$

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 $a = x^{4}$ and $b = y^{3}$ .

$(x^{4} - y^{3}) ({(x^{4})}^{2} + x^{4} y^{3} + {(y^{3})}^{2})$

Simplify.

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

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

$(x^{4} - y^{3}) (x^{4 \cdot 2} + x^{4} y^{3} + {(y^{3})}^{2})$

Multiply $4$ by $2$ .

$(x^{4} - y^{3}) (x^{8} + x^{4} y^{3} + {(y^{3})}^{2})$

Multiply the exponents in ${(y^{3})}^{2}$ .

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

$(x^{4} - y^{3}) (x^{8} + x^{4} y^{3} + y^{3 \cdot 2})$

Multiply $3$ by $2$ .

$(x^{4} - y^{3}) (x^{8} + x^{4} y^{3} + y^{6})$

Factor x^12-y^9

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