Factor 27x^3+64y^3

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

Step 1

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

${(3 x)}^{3} + 64 y^{3}$

Step 2

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

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

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 $a = 3 x$ and $b = 4 y$ .

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

Step 4

Simplify.

Apply the product rule to $3 x$ .

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

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

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

Multiply $3$ by $- 1$ .

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

Rewrite using the commutative property of multiplication.

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

Multiply $- 3$ by $4$ .

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

Apply the product rule to $4 y$ .

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

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

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

Factor 27x^3+64y^3

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