Algebra Examples

$1 - 64 p^{6}$

Step 1

Rewrite $1$ as $1^{3}$ .

$1^{3} - 64 p^{6}$

Step 2

Rewrite $64 p^{6}$ as ${(4 p^{2})}^{3}$ .

$1^{3} - {(4 p^{2})}^{3}$

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 $a = 1$ and $b = 4 p^{2}$ .

$(1 - (4 p^{2})) (1^{2} + 1 (4 p^{2}) + {(4 p^{2})}^{2})$

Step 4

Simplify.

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Step 4.1

Remove parentheses.

$(1 - 1 \cdot 4 p^{2}) (1^{2} + 1 (4 p^{2}) + {(4 p^{2})}^{2})$

Step 4.2

Rewrite $1$ as $1^{2}$ .

$(1^{2} - 1 \cdot 4 p^{2}) (1^{2} + 1 (4 p^{2}) + {(4 p^{2})}^{2})$

Step 4.3

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

$(1^{2} - {(2 p)}^{2}) (1^{2} + 1 (4 p^{2}) + {(4 p^{2})}^{2})$

Step 4.4

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1$ and $b = 2 p$ .

$(1 + 2 p) (1 - (2 p)) (1^{2} + 1 (4 p^{2}) + {(4 p^{2})}^{2})$

Step 4.5

Multiply $2$ by $- 1$ .

$(1 + 2 p) (1 - 2 p) (1^{2} + 1 (4 p^{2}) + {(4 p^{2})}^{2})$

Step 4.6

Multiply $4$ by $1$ .

$(1 + 2 p) (1 - 2 p) (1^{2} + 4 p^{2} + {(4 p^{2})}^{2})$

Step 5

One to any power is one.

$(1 + 2 p) (1 - 2 p) (1 + 4 p^{2} + {(4 p^{2})}^{2})$

Step 6

Apply the product rule to $4 p^{2}$ .

$(1 + 2 p) (1 - 2 p) (1 + 4 p^{2} + 4^{2} {(p^{2})}^{2})$

Step 7

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

$(1 + 2 p) (1 - 2 p) (1 + 4 p^{2} + 16 {(p^{2})}^{2})$

Step 8

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

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Step 8.1

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

$(1 + 2 p) (1 - 2 p) (1 + 4 p^{2} + 16 p^{2 \cdot 2})$

Step 8.2

Multiply $2$ by $2$ .

$(1 + 2 p) (1 - 2 p) (1 + 4 p^{2} + 16 p^{4})$

Step 9

Factor.

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Step 9.1

Rewrite $1 + 4 p^{2} + 16 p^{4}$ in a factored form.

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Step 9.1.1

Rewrite the middle term.

$(1 + 2 p) (1 - 2 p) (1 + 2 \cdot 1 (4 p^{2}) - 4 p^{2} + 16 p^{4})$

Step 9.1.2

Rearrange terms.

$(1 + 2 p) (1 - 2 p) (1 + 2 \cdot 1 (4 p^{2}) + 16 p^{4} - 4 p^{2})$

Step 9.1.3

Factor first three terms by perfect square rule.

$(1 + 2 p) (1 - 2 p) ({(1 + 4 p^{2})}^{2} - 4 p^{2})$

Step 9.1.4

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

$(1 + 2 p) (1 - 2 p) ({(1 + 4 p^{2})}^{2} - {(2 p)}^{2})$

Step 9.1.5

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 1 + 4 p^{2}$ and $b = 2 p$ .

$(1 + 2 p) (1 - 2 p) ((1 + 4 p^{2} + 2 p) (1 + 4 p^{2} - (2 p)))$

Step 9.1.6

Simplify.

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Step 9.1.6.1

Reorder terms.

$(1 + 2 p) (1 - 2 p) ((4 p^{2} + 2 p + 1) (1 + 4 p^{2} - (2 p)))$

Step 9.1.6.2

Multiply $2$ by $- 1$ .

$(1 + 2 p) (1 - 2 p) ((4 p^{2} + 2 p + 1) (1 + 4 p^{2} - 2 p))$

Step 9.1.6.3

Reorder terms.

$(1 + 2 p) (1 - 2 p) ((4 p^{2} + 2 p + 1) (4 p^{2} - 2 p + 1))$

Step 9.2

Remove unnecessary parentheses.

$(1 + 2 p) (1 - 2 p) (4 p^{2} + 2 p + 1) (4 p^{2} - 2 p + 1)$