Finite Math Examples

$216 p^{3} - 125 q^{3}$

Step 1

Rewrite $216 p^{3}$ as ${(6 p)}^{3}$ .

${(6 p)}^{3} - 125 q^{3}$

Step 2

Rewrite $125 q^{3}$ as ${(5 q)}^{3}$ .

${(6 p)}^{3} - {(5 q)}^{3}$

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 $a = 6 p$ and $b = 5 q$ .

$(6 p - (5 q)) ({(6 p)}^{2} + 6 p (5 q) + {(5 q)}^{2})$

Step 4

Simplify.

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Step 4.1

Multiply $5$ by $- 1$ .

$(6 p - 5 q) ({(6 p)}^{2} + 6 p (5 q) + {(5 q)}^{2})$

Step 4.2

Apply the product rule to $6 p$ .

$(6 p - 5 q) (6^{2} p^{2} + 6 p (5 q) + {(5 q)}^{2})$

Step 4.3

Raise $6$ to the power of $2$ .

$(6 p - 5 q) (36 p^{2} + 6 p (5 q) + {(5 q)}^{2})$

Step 4.4

Rewrite using the commutative property of multiplication.

$(6 p - 5 q) (36 p^{2} + 6 \cdot 5 p q + {(5 q)}^{2})$

Step 4.5

Multiply $6$ by $5$ .

$(6 p - 5 q) (36 p^{2} + 30 p q + {(5 q)}^{2})$

Step 4.6

Apply the product rule to $5 q$ .

$(6 p - 5 q) (36 p^{2} + 30 p q + 5^{2} q^{2})$

Step 4.7

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

$(6 p - 5 q) (36 p^{2} + 30 p q + 25 q^{2})$