Factor 125a^3+150a^2b+60ab^2+8b^3

$125 a^{3} + 150 a^{2} b + 60 a b^{2} + 8 b^{3}$

Step 1

Regroup terms.

$125 a^{3} + 8 b^{3} + 150 a^{2} b + 60 a b^{2}$

Step 2

Rewrite $125 a^{3}$ as ${(5 a)}^{3}$ .

${(5 a)}^{3} + 8 b^{3} + 150 a^{2} b + 60 a b^{2}$

Step 3

Rewrite $8 b^{3}$ as ${(2 b)}^{3}$ .

${(5 a)}^{3} + {(2 b)}^{3} + 150 a^{2} b + 60 a b^{2}$

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 $a = 5 a$ and $b = 2 b$ .

$(5 a + 2 b) ({(5 a)}^{2} - (5 a) (2 b) + {(2 b)}^{2}) + 150 a^{2} b + 60 a b^{2}$

Step 5

Simplify.

Apply the product rule to $5 a$ .

$(5 a + 2 b) (5^{2} a^{2} - (5 a) (2 b) + {(2 b)}^{2}) + 150 a^{2} b + 60 a b^{2}$

Raise $5$ to the power of $2$ .

$(5 a + 2 b) (25 a^{2} - (5 a) (2 b) + {(2 b)}^{2}) + 150 a^{2} b + 60 a b^{2}$

Multiply $5$ by $- 1$ .

$(5 a + 2 b) (25 a^{2} - 5 a (2 b) + {(2 b)}^{2}) + 150 a^{2} b + 60 a b^{2}$

Rewrite using the commutative property of multiplication.

$(5 a + 2 b) (25 a^{2} - 5 \cdot 2 a b + {(2 b)}^{2}) + 150 a^{2} b + 60 a b^{2}$

Multiply $- 5$ by $2$ .

$(5 a + 2 b) (25 a^{2} - 10 a b + {(2 b)}^{2}) + 150 a^{2} b + 60 a b^{2}$

Apply the product rule to $2 b$ .

$(5 a + 2 b) (25 a^{2} - 10 a b + 2^{2} b^{2}) + 150 a^{2} b + 60 a b^{2}$

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

$(5 a + 2 b) (25 a^{2} - 10 a b + 4 b^{2}) + 150 a^{2} b + 60 a b^{2}$

Step 6

Factor $30 a b$ out of $150 a^{2} b + 60 a b^{2}$ .

Factor $30 a b$ out of $150 a^{2} b$ .

$(5 a + 2 b) (25 a^{2} - 10 a b + 4 b^{2}) + 30 a b (5 a) + 60 a b^{2}$

Factor $30 a b$ out of $60 a b^{2}$ .

$(5 a + 2 b) (25 a^{2} - 10 a b + 4 b^{2}) + 30 a b (5 a) + 30 a b (2 b)$

Factor $30 a b$ out of $30 a b (5 a) + 30 a b (2 b)$ .

$(5 a + 2 b) (25 a^{2} - 10 a b + 4 b^{2}) + 30 a b (5 a + 2 b)$

Step 7

Factor $5 a + 2 b$ out of $(5 a + 2 b) (25 a^{2} - 10 a b + 4 b^{2}) + 30 a b (5 a + 2 b)$ .

Factor $5 a + 2 b$ out of $30 a b (5 a + 2 b)$ .

$(5 a + 2 b) (25 a^{2} - 10 a b + 4 b^{2}) + (5 a + 2 b) (30 a b)$

Factor $5 a + 2 b$ out of $(5 a + 2 b) (25 a^{2} - 10 a b + 4 b^{2}) + (5 a + 2 b) (30 a b)$ .

$(5 a + 2 b) (25 a^{2} - 10 a b + 4 b^{2} + 30 a b)$

Step 8

Add $- 10 a b$ and $30 a b$ .

$(5 a + 2 b) (25 a^{2} + 4 b^{2} + 20 a b)$

Step 9

Factor using the perfect square rule.

Rearrange terms.

$(5 a + 2 b) (25 a^{2} + 20 a b + 4 b^{2})$

Rewrite $25 a^{2}$ as ${(5 a)}^{2}$ .

$(5 a + 2 b) ({(5 a)}^{2} + 20 a b + 4 b^{2})$

Rewrite $4 b^{2}$ as ${(2 b)}^{2}$ .

$(5 a + 2 b) ({(5 a)}^{2} + 20 a b + {(2 b)}^{2})$

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

$20 a b = 2 \cdot (5 a) \cdot (2 b)$

Rewrite the polynomial.

$(5 a + 2 b) ({(5 a)}^{2} + 2 \cdot (5 a) \cdot (2 b) + {(2 b)}^{2})$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 5 a$ and $b = 2 b$ .

$(5 a + 2 b) {(5 a + 2 b)}^{2}$

Factor 125a^3+150a^2b+60ab^2+8b^3

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