Factor x^6-27y^3

$x^{6} - 27 y^{3}$

Step 1

Rewrite $x^{6}$ as ${(x^{2})}^{3}$ .

${(x^{2})}^{3} - 27 y^{3}$

Step 2

Rewrite $27 y^{3}$ as ${(3 y)}^{3}$ .

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

Step 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 $a = x^{2}$ and $b = 3 y$ .

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

Step 4

Simplify.

Multiply $3$ by $- 1$ .

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

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

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

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

Multiply $2$ by $2$ .

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

Rewrite using the commutative property of multiplication.

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

Apply the product rule to $3 y$ .

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

Raise $3$ to the power of $2$ .

$(x^{2} - 3 y) (x^{4} + 3 x^{2} y + 9 y^{2})$

Factor x^6-27y^3

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