Factor 27m^3+108m^2n+144mn^2+64n^3

$27 m^{3} + 108 m^{2} n + 144 m n^{2} + 64 n^{3}$

Step 1

Regroup terms.

$27 m^{3} + 64 n^{3} + 108 m^{2} n + 144 m n^{2}$

Step 2

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

${(3 m)}^{3} + 64 n^{3} + 108 m^{2} n + 144 m n^{2}$

Step 3

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

${(3 m)}^{3} + {(4 n)}^{3} + 108 m^{2} n + 144 m n^{2}$

Step 4

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 $a = 3 m$ and $b = 4 n$ .

$(3 m + 4 n) ({(3 m)}^{2} - (3 m) (4 n) + {(4 n)}^{2}) + 108 m^{2} n + 144 m n^{2}$

Step 5

Simplify.

Apply the product rule to $3 m$ .

$(3 m + 4 n) (3^{2} m^{2} - (3 m) (4 n) + {(4 n)}^{2}) + 108 m^{2} n + 144 m n^{2}$

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

$(3 m + 4 n) (9 m^{2} - (3 m) (4 n) + {(4 n)}^{2}) + 108 m^{2} n + 144 m n^{2}$

Multiply $3$ by $- 1$ .

$(3 m + 4 n) (9 m^{2} - 3 m (4 n) + {(4 n)}^{2}) + 108 m^{2} n + 144 m n^{2}$

Rewrite using the commutative property of multiplication.

$(3 m + 4 n) (9 m^{2} - 3 \cdot 4 m n + {(4 n)}^{2}) + 108 m^{2} n + 144 m n^{2}$

Multiply $- 3$ by $4$ .

$(3 m + 4 n) (9 m^{2} - 12 m n + {(4 n)}^{2}) + 108 m^{2} n + 144 m n^{2}$

Apply the product rule to $4 n$ .

$(3 m + 4 n) (9 m^{2} - 12 m n + 4^{2} n^{2}) + 108 m^{2} n + 144 m n^{2}$

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

$(3 m + 4 n) (9 m^{2} - 12 m n + 16 n^{2}) + 108 m^{2} n + 144 m n^{2}$

Step 6

Factor $36 m n$ out of $108 m^{2} n + 144 m n^{2}$ .

Factor $36 m n$ out of $108 m^{2} n$ .

$(3 m + 4 n) (9 m^{2} - 12 m n + 16 n^{2}) + 36 m n (3 m) + 144 m n^{2}$

Factor $36 m n$ out of $144 m n^{2}$ .

$(3 m + 4 n) (9 m^{2} - 12 m n + 16 n^{2}) + 36 m n (3 m) + 36 m n (4 n)$

Factor $36 m n$ out of $36 m n (3 m) + 36 m n (4 n)$ .

$(3 m + 4 n) (9 m^{2} - 12 m n + 16 n^{2}) + 36 m n (3 m + 4 n)$

Step 7

Factor $3 m + 4 n$ out of $(3 m + 4 n) (9 m^{2} - 12 m n + 16 n^{2}) + 36 m n (3 m + 4 n)$ .

Factor $3 m + 4 n$ out of $36 m n (3 m + 4 n)$ .

$(3 m + 4 n) (9 m^{2} - 12 m n + 16 n^{2}) + (3 m + 4 n) (36 m n)$

Factor $3 m + 4 n$ out of $(3 m + 4 n) (9 m^{2} - 12 m n + 16 n^{2}) + (3 m + 4 n) (36 m n)$ .

$(3 m + 4 n) (9 m^{2} - 12 m n + 16 n^{2} + 36 m n)$

Step 8

Add $- 12 m n$ and $36 m n$ .

$(3 m + 4 n) (9 m^{2} + 16 n^{2} + 24 m n)$

Step 9

Factor using the perfect square rule.

Rearrange terms.

$(3 m + 4 n) (9 m^{2} + 24 m n + 16 n^{2})$

Rewrite $9 m^{2}$ as ${(3 m)}^{2}$ .

$(3 m + 4 n) ({(3 m)}^{2} + 24 m n + 16 n^{2})$

Rewrite $16 n^{2}$ as ${(4 n)}^{2}$ .

$(3 m + 4 n) ({(3 m)}^{2} + 24 m n + {(4 n)}^{2})$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$24 m n = 2 \cdot (3 m) \cdot (4 n)$

Rewrite the polynomial.

$(3 m + 4 n) ({(3 m)}^{2} + 2 \cdot (3 m) \cdot (4 n) + {(4 n)}^{2})$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 3 m$ and $b = 4 n$ .

$(3 m + 4 n) {(3 m + 4 n)}^{2}$

Factor 27m^3+108m^2n+144mn^2+64n^3

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