Pre-Algebra Examples

$\frac{64 x^{3} - 27}{4 x - 3}$

Step 1

Simplify the numerator.

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Step 1.1

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

$\frac{{(4 x)}^{3} - 27}{4 x - 3}$

Step 1.2

Rewrite $27$ as $3^{3}$ .

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

Step 1.3

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 $b = 3$ .

$\frac{(4 x - 3) ({(4 x)}^{2} + 4 x \cdot 3 + 3^{2})}{4 x - 3}$

Step 1.4

Simplify.

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Step 1.4.1

Apply the product rule to $4 x$ .

$\frac{(4 x - 3) (4^{2} x^{2} + 4 x \cdot 3 + 3^{2})}{4 x - 3}$

Step 1.4.2

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

$\frac{(4 x - 3) (16 x^{2} + 4 x \cdot 3 + 3^{2})}{4 x - 3}$

Step 1.4.3

Multiply $3$ by $4$ .

$\frac{(4 x - 3) (16 x^{2} + 12 x + 3^{2})}{4 x - 3}$

Step 1.4.4

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

$\frac{(4 x - 3) (16 x^{2} + 12 x + 9)}{4 x - 3}$

Step 2

Cancel the common factor of $4 x - 3$ .

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Step 2.1

Cancel the common factor.

$\frac{(4 x - 3) (16 x^{2} + 12 x + 9)}{4 x - 3}$

Step 2.2

Divide $16 x^{2} + 12 x + 9$ by $1$ .

$16 x^{2} + 12 x + 9$