Factor m^3+64n^3

$m^{3} + 64 n^{3}$

Step 1

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

$m^{3} + {(4 n)}^{3}$

Step 2

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 $a = m$ and $b = 4 n$ .

$(m + 4 n) (m^{2} - m (4 n) + {(4 n)}^{2})$

Step 3

Simplify.

Rewrite using the commutative property of multiplication.

$(m + 4 n) (m^{2} - 1 \cdot 4 m n + {(4 n)}^{2})$

Multiply $- 1$ by $4$ .

$(m + 4 n) (m^{2} - 4 m n + {(4 n)}^{2})$

Apply the product rule to $4 n$ .

$(m + 4 n) (m^{2} - 4 m n + 4^{2} n^{2})$

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

$(m + 4 n) (m^{2} - 4 m n + 16 n^{2})$

Factor m^3+64n^3

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