Pre-Algebra Examples

$8 a^{4} - 125 a b^{3}$

Step 1

Factor $a$ out of $8 a^{4} - 125 a b^{3}$ .

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Step 1.1

Factor $a$ out of $8 a^{4}$ .

$a (8 a^{3}) - 125 a b^{3}$

Step 1.2

Factor $a$ out of $- 125 a b^{3}$ .

$a (8 a^{3}) + a (- 125 b^{3})$

Step 1.3

Factor $a$ out of $a (8 a^{3}) + a (- 125 b^{3})$ .

$a (8 a^{3} - 125 b^{3})$

Step 2

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

$a ({(2 a)}^{3} - 125 b^{3})$

Step 3

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

$a ({(2 a)}^{3} - {(5 b)}^{3})$

Step 4

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = 2 a$ and $b = 5 b$ .

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

Step 5

Factor.

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Step 5.1

Simplify.

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Step 5.1.1

Multiply $5$ by $- 1$ .

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

Step 5.1.2

Apply the product rule to $2 a$ .

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

Step 5.1.3

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

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

Step 5.1.4

Rewrite using the commutative property of multiplication.

$a ((2 a - 5 b) (4 a^{2} + 2 \cdot 5 a b + {(5 b)}^{2}))$

Step 5.1.5

Multiply $2$ by $5$ .

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

Step 5.1.6

Apply the product rule to $5 b$ .

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

Step 5.1.7

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

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

Step 5.2

Remove unnecessary parentheses.

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