Factor 10+270y^3

$10 + 270 y^{3}$

Factor $10$ out of $10 + 270 y^{3}$ .

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Factor $10$ out of $10$ .

$10 (1) + 270 y^{3}$

Factor $10$ out of $270 y^{3}$ .

$10 (1) + 10 (27 y^{3})$

Factor $10$ out of $10 (1) + 10 (27 y^{3})$ .

$10 (1 + 27 y^{3})$

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

$10 (1^{3} + 27 y^{3})$

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

$10 (1^{3} + {(3 y)}^{3})$

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

$10 ((1 + 3 y) (1^{2} - 1 (3 y) + {(3 y)}^{2}))$

Factor.

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Simplify.

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One to any power is one.

$10 ((1 + 3 y) (1 - 1 (3 y) + {(3 y)}^{2}))$

Multiply $3$ by $- 1$ .

$10 ((1 + 3 y) (1 - 3 y + {(3 y)}^{2}))$

Apply the product rule to $3 y$ .

$10 ((1 + 3 y) (1 - 3 y + 3^{2} y^{2}))$

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

$10 ((1 + 3 y) (1 - 3 y + 9 y^{2}))$

Remove unnecessary parentheses.

$10 (1 + 3 y) (1 - 3 y + 9 y^{2})$