Algebra Examples

$8 a^{3} - 64 b^{6}$

Step 1

Factor out the GCF of $8$ from $8 a^{3} - 64 b^{6}$ .

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Step 1.1

Factor out the GCF of $8$ from each term in the polynomial.

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Step 1.1.1

Factor out the GCF of $8$ from the expression $8 a^{3}$ .

$8 (a^{3}) - 64 b^{6}$

Step 1.1.2

Factor out the GCF of $8$ from the expression $- 64 b^{6}$ .

$8 (a^{3}) + 8 (- 8 b^{6})$

Step 1.2

Since all the terms share a common factor of $8$ , it can be factored out of each term.

$8 (a^{3} - 8 b^{6})$

Step 2

Rewrite $8 b^{6}$ as ${(2 b^{2})}^{3}$ .

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

Step 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 $a = a$ and $b = 2 b^{2}$ .

$8 ((a - (2 b^{2})) (a^{2} + a (2 b^{2}) + {(2 b^{2})}^{2}))$

Step 4

Simplify.

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Step 4.1

Multiply $2$ by $- 1$ .

$8 ((a - 2 b^{2}) (a^{2} + a (2 b^{2}) + {(2 b^{2})}^{2}))$

Step 4.2

Rewrite using the commutative property of multiplication.

$8 ((a - 2 b^{2}) (a^{2} + 2 a b^{2} + {(2 b^{2})}^{2}))$

Step 4.3

Apply the product rule to $2 b^{2}$ .

$8 ((a - 2 b^{2}) (a^{2} + 2 a b^{2} + 2^{2} {(b^{2})}^{2}))$

Step 4.4

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

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

Step 4.5

Multiply the exponents in ${(b^{2})}^{2}$ .

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Step 4.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 4.5.2

Multiply $2$ by $2$ .

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