Factor 27t^3+8

$27 t^{3} + 8$

Step 1

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

${(3 t)}^{3} + 8$

Step 2

Rewrite $8$ as $2^{3}$ .

${(3 t)}^{3} + 2^{3}$

Step 3

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 3 t$ and $b = 2$ .

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

Step 4

Simplify.

Apply the product rule to $3 t$ .

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

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

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

Multiply $3$ by $- 1$ .

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

Multiply $2$ by $- 3$ .

$(3 t + 2) (9 t^{2} - 6 t + 2^{2})$

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

$(3 t + 2) (9 t^{2} - 6 t + 4)$

Factor 27t^3+8

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