Pre-Algebra Examples

$64 a^{3} + 20 a - 125 b^{3} - 25 b$

Step 1

Regroup terms.

$64 a^{3} - 125 b^{3} + 20 a - 25 b$

Step 2

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

${(4 a)}^{3} - 125 b^{3} + 20 a - 25 b$

Step 3

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

${(4 a)}^{3} - {(5 b)}^{3} + 20 a - 25 b$

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 $a = 4 a$ and $b = 5 b$ .

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

Step 5

Simplify.

Tap for more steps...

Step 5.1

Multiply $5$ by $- 1$ .

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

Step 5.2

Apply the product rule to $4 a$ .

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

Step 5.3

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

$(4 a - 5 b) (16 a^{2} + 4 a (5 b) + {(5 b)}^{2}) + 20 a - 25 b$

Step 5.4

Rewrite using the commutative property of multiplication.

$(4 a - 5 b) (16 a^{2} + 4 \cdot 5 a b + {(5 b)}^{2}) + 20 a - 25 b$

Step 5.5

Multiply $4$ by $5$ .

$(4 a - 5 b) (16 a^{2} + 20 a b + {(5 b)}^{2}) + 20 a - 25 b$

Step 5.6

Apply the product rule to $5 b$ .

$(4 a - 5 b) (16 a^{2} + 20 a b + 5^{2} b^{2}) + 20 a - 25 b$

Step 5.7

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

$(4 a - 5 b) (16 a^{2} + 20 a b + 25 b^{2}) + 20 a - 25 b$

Step 6

Factor $5$ out of $20 a - 25 b$ .

Tap for more steps...

Step 6.1

Factor $5$ out of $20 a$ .

$(4 a - 5 b) (16 a^{2} + 20 a b + 25 b^{2}) + 5 (4 a) - 25 b$

Step 6.2

Factor $5$ out of $- 25 b$ .

$(4 a - 5 b) (16 a^{2} + 20 a b + 25 b^{2}) + 5 (4 a) + 5 (- 5 b)$

Step 6.3

Factor $5$ out of $5 (4 a) + 5 (- 5 b)$ .

$(4 a - 5 b) (16 a^{2} + 20 a b + 25 b^{2}) + 5 (4 a - 5 b)$

Step 7

Factor $4 a - 5 b$ out of $(4 a - 5 b) (16 a^{2} + 20 a b + 25 b^{2}) + 5 (4 a - 5 b)$ .

Tap for more steps...

Step 7.1

Factor $4 a - 5 b$ out of $5 (4 a - 5 b)$ .

$(4 a - 5 b) (16 a^{2} + 20 a b + 25 b^{2}) + (4 a - 5 b) \cdot 5$

Step 7.2

Factor $4 a - 5 b$ out of $(4 a - 5 b) (16 a^{2} + 20 a b + 25 b^{2}) + (4 a - 5 b) \cdot 5$ .

$(4 a - 5 b) (16 a^{2} + 20 a b + 25 b^{2} + 5)$