Pre-Algebra Examples

$64 x^{3} - 125 a^{6}$

Step 1

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

${(4 x)}^{3} - 125 a^{6}$

Step 2

Rewrite $125 a^{6}$ as ${(5 a^{2})}^{3}$ .

${(4 x)}^{3} - {(5 a^{2})}^{3}$

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 $a = 4 x$ and $b = 5 a^{2}$ .

$(4 x - (5 a^{2})) ({(4 x)}^{2} + 4 x (5 a^{2}) + {(5 a^{2})}^{2})$

Step 4

Simplify.

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Step 4.1

Multiply $5$ by $- 1$ .

$(4 x - 5 a^{2}) ({(4 x)}^{2} + 4 x (5 a^{2}) + {(5 a^{2})}^{2})$

Step 4.2

Apply the product rule to $4 x$ .

$(4 x - 5 a^{2}) (4^{2} x^{2} + 4 x (5 a^{2}) + {(5 a^{2})}^{2})$

Step 4.3

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

$(4 x - 5 a^{2}) (16 x^{2} + 4 x (5 a^{2}) + {(5 a^{2})}^{2})$

Step 4.4

Rewrite using the commutative property of multiplication.

$(4 x - 5 a^{2}) (16 x^{2} + 4 \cdot 5 x a^{2} + {(5 a^{2})}^{2})$

Step 4.5

Multiply $4$ by $5$ .

$(4 x - 5 a^{2}) (16 x^{2} + 20 x a^{2} + {(5 a^{2})}^{2})$

Step 4.6

Apply the product rule to $5 a^{2}$ .

$(4 x - 5 a^{2}) (16 x^{2} + 20 x a^{2} + 5^{2} {(a^{2})}^{2})$

Step 4.7

Raise $5$ to the power of $2$ .

$(4 x - 5 a^{2}) (16 x^{2} + 20 x a^{2} + 25 {(a^{2})}^{2})$

Step 4.8

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

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Step 4.8.1

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

$(4 x - 5 a^{2}) (16 x^{2} + 20 x a^{2} + 25 a^{2 \cdot 2})$

Step 4.8.2

Multiply $2$ by $2$ .

$(4 x - 5 a^{2}) (16 x^{2} + 20 x a^{2} + 25 a^{4})$