Pre-Algebra Examples

$\frac{49 x^{2} - 28 x + 16}{343 x^{3} + 64}$

Step 1

Simplify the denominator.

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Step 1.1

Rewrite $343 x^{3}$ as ${(7 x)}^{3}$ .

$\frac{49 x^{2} - 28 x + 16}{{(7 x)}^{3} + 64}$

Step 1.2

Rewrite $64$ as $4^{3}$ .

$\frac{49 x^{2} - 28 x + 16}{{(7 x)}^{3} + 4^{3}}$

Step 1.3

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 $a = 7 x$ and $b = 4$ .

$\frac{49 x^{2} - 28 x + 16}{(7 x + 4) ({(7 x)}^{2} - (7 x) \cdot 4 + 4^{2})}$

Step 1.4

Simplify.

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Step 1.4.1

Apply the product rule to $7 x$ .

$\frac{49 x^{2} - 28 x + 16}{(7 x + 4) (7^{2} x^{2} - (7 x) \cdot 4 + 4^{2})}$

Step 1.4.2

Raise $7$ to the power of $2$ .

$\frac{49 x^{2} - 28 x + 16}{(7 x + 4) (49 x^{2} - (7 x) \cdot 4 + 4^{2})}$

Step 1.4.3

Multiply $7$ by $- 1$ .

$\frac{49 x^{2} - 28 x + 16}{(7 x + 4) (49 x^{2} - 7 x \cdot 4 + 4^{2})}$

Step 1.4.4

Multiply $4$ by $- 7$ .

$\frac{49 x^{2} - 28 x + 16}{(7 x + 4) (49 x^{2} - 28 x + 4^{2})}$

Step 1.4.5

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

$\frac{49 x^{2} - 28 x + 16}{(7 x + 4) (49 x^{2} - 28 x + 16)}$

Step 2

Cancel the common factor of $49 x^{2} - 28 x + 16$ .

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Step 2.1

Cancel the common factor.

$\frac{49 x^{2} - 28 x + 16}{(7 x + 4) (49 x^{2} - 28 x + 16)}$

Step 2.2

Rewrite the expression.

$\frac{1}{7 x + 4}$