Pre-Algebra Examples

$(64 x^{3} - 125) \div (4 x - 5)$

Step 1

Rewrite the division as a fraction.

$\frac{64 x^{3} - 125}{4 x - 5}$

Step 2

Simplify the numerator.

Tap for more steps...

Step 2.1

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

$\frac{{(4 x)}^{3} - 125}{4 x - 5}$

Step 2.2

Rewrite $125$ as $5^{3}$ .

$\frac{{(4 x)}^{3} - 5^{3}}{4 x - 5}$

Step 2.3

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

$\frac{(4 x - 5) ({(4 x)}^{2} + 4 x \cdot 5 + 5^{2})}{4 x - 5}$

Step 2.4

Simplify.

Tap for more steps...

Step 2.4.1

Apply the product rule to $4 x$ .

$\frac{(4 x - 5) (4^{2} x^{2} + 4 x \cdot 5 + 5^{2})}{4 x - 5}$

Step 2.4.2

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

$\frac{(4 x - 5) (16 x^{2} + 4 x \cdot 5 + 5^{2})}{4 x - 5}$

Step 2.4.3

Multiply $5$ by $4$ .

$\frac{(4 x - 5) (16 x^{2} + 20 x + 5^{2})}{4 x - 5}$

Step 2.4.4

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

$\frac{(4 x - 5) (16 x^{2} + 20 x + 25)}{4 x - 5}$

Step 3

Cancel the common factor of $4 x - 5$ .

Tap for more steps...

Step 3.1

Cancel the common factor.

$\frac{(4 x - 5) (16 x^{2} + 20 x + 25)}{4 x - 5}$

Step 3.2

Divide $16 x^{2} + 20 x + 25$ by $1$ .

$16 x^{2} + 20 x + 25$