Precalculus Examples

$27 s^{3} - 125 t^{3}$

Step 1

Rewrite $27 s^{3}$ as ${(3 s)}^{3}$ .

${(3 s)}^{3} - 125 t^{3}$

Step 2

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

${(3 s)}^{3} - {(5 t)}^{3}$

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 $a = 3 s$ and $b = 5 t$ .

$(3 s - (5 t)) ({(3 s)}^{2} + 3 s (5 t) + {(5 t)}^{2})$

Step 4

Simplify.

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Step 4.1

Multiply $5$ by $- 1$ .

$(3 s - 5 t) ({(3 s)}^{2} + 3 s (5 t) + {(5 t)}^{2})$

Step 4.2

Apply the product rule to $3 s$ .

$(3 s - 5 t) (3^{2} s^{2} + 3 s (5 t) + {(5 t)}^{2})$

Step 4.3

Raise $3$ to the power of $2$ .

$(3 s - 5 t) (9 s^{2} + 3 s (5 t) + {(5 t)}^{2})$

Step 4.4

Rewrite using the commutative property of multiplication.

$(3 s - 5 t) (9 s^{2} + 3 \cdot 5 s t + {(5 t)}^{2})$

Step 4.5

Multiply $3$ by $5$ .

$(3 s - 5 t) (9 s^{2} + 15 s t + {(5 t)}^{2})$

Step 4.6

Apply the product rule to $5 t$ .

$(3 s - 5 t) (9 s^{2} + 15 s t + 5^{2} t^{2})$

Step 4.7

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

$(3 s - 5 t) (9 s^{2} + 15 s t + 25 t^{2})$