Factor 64x^3+27y^3

$64 x^{3} + 27 y^{3}$

Step 1

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

${(4 x)}^{3} + 27 y^{3}$

Step 2

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

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

Step 3

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

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

Step 4

Simplify.

Apply the product rule to $4 x$ .

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

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

$(4 x + 3 y) (16 x^{2} - (4 x) (3 y) + {(3 y)}^{2})$

Multiply $4$ by $- 1$ .

$(4 x + 3 y) (16 x^{2} - 4 x (3 y) + {(3 y)}^{2})$

Rewrite using the commutative property of multiplication.

$(4 x + 3 y) (16 x^{2} - 4 \cdot 3 x y + {(3 y)}^{2})$

Multiply $- 4$ by $3$ .

$(4 x + 3 y) (16 x^{2} - 12 x y + {(3 y)}^{2})$

Apply the product rule to $3 y$ .

$(4 x + 3 y) (16 x^{2} - 12 x y + 3^{2} y^{2})$

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

$(4 x + 3 y) (16 x^{2} - 12 x y + 9 y^{2})$

Factor 64x^3+27y^3

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