Algebra Examples

$\frac{a^{2} - 4 a b + 4 b^{2}}{a^{3} - 8 b^{3}}$

Step 1

Factor using the perfect square rule.

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Step 1.1

Rewrite $4 b^{2}$ as ${(2 b)}^{2}$ .

$\frac{a^{2} - 4 a b + {(2 b)}^{2}}{a^{3} - 8 b^{3}}$

Step 1.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$4 a b = 2 \cdot a \cdot (2 b)$

Step 1.3

Rewrite the polynomial.

$\frac{a^{2} - 2 \cdot a \cdot (2 b) + {(2 b)}^{2}}{a^{3} - 8 b^{3}}$

Step 1.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = a$ and $b = 2 b$ .

$\frac{{(a - 2 b)}^{2}}{a^{3} - 8 b^{3}}$

Step 2

Simplify the denominator.

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Step 2.1

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

$\frac{{(a - 2 b)}^{2}}{a^{3} - {(2 b)}^{3}}$

Step 2.2

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 $a = a$ and $b = 2 b$ .

$\frac{{(a - 2 b)}^{2}}{(a - (2 b)) (a^{2} + a (2 b) + {(2 b)}^{2})}$

Step 2.3

Simplify.

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Step 2.3.1

Multiply $2$ by $- 1$ .

$\frac{{(a - 2 b)}^{2}}{(a - 2 b) (a^{2} + a (2 b) + {(2 b)}^{2})}$

Step 2.3.2

Rewrite using the commutative property of multiplication.

$\frac{{(a - 2 b)}^{2}}{(a - 2 b) (a^{2} + 2 a b + {(2 b)}^{2})}$

Step 2.3.3

Apply the product rule to $2 b$ .

$\frac{{(a - 2 b)}^{2}}{(a - 2 b) (a^{2} + 2 a b + 2^{2} b^{2})}$

Step 2.3.4

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

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

Step 3

Cancel the common factor of ${(a - 2 b)}^{2}$ and $a - 2 b$ .

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Step 3.1

Factor $a - 2 b$ out of ${(a - 2 b)}^{2}$ .

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

Step 3.2

Cancel the common factors.

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Step 3.2.1

Cancel the common factor.

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

Step 3.2.2

Rewrite the expression.

$\frac{a - 2 b}{a^{2} + 2 a b + 4 b^{2}}$