Factor x^6-64y^3

$x^{6} - 64 y^{3}$

Step 1

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

${(x^{2})}^{3} - 64 y^{3}$

Step 2

Rewrite $64 y^{3}$ as ${(4 y)}^{3}$ .

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

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 $b = 4 y$ .

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

Step 4

Simplify.

Multiply $4$ by $- 1$ .

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

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

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

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

Multiply $2$ by $2$ .

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

Rewrite using the commutative property of multiplication.

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

Apply the product rule to $4 y$ .

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

Raise $4$ to the power of $2$ .

$(x^{2} - 4 y) (x^{4} + 4 x^{2} y + 16 y^{2})$

Factor x^6-64y^3

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