Factor 27y^3-1

$27 y^{3} - 1$

Step 1

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

${(3 y)}^{3} - 1$

Step 2

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

${(3 y)}^{3} - 1^{3}$

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 $a = 3 y$ and $b = 1$ .

$(3 y - 1) ({(3 y)}^{2} + 3 y \cdot 1 + 1^{2})$

Step 4

Simplify.

Apply the product rule to $3 y$ .

$(3 y - 1) (3^{2} y^{2} + 3 y \cdot 1 + 1^{2})$

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

$(3 y - 1) (9 y^{2} + 3 y \cdot 1 + 1^{2})$

Multiply $3$ by $1$ .

$(3 y - 1) (9 y^{2} + 3 y + 1^{2})$

One to any power is one.

$(3 y - 1) (9 y^{2} + 3 y + 1)$

Factor 27y^3-1

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