Factor a^(3y)+1

$a^{3 y} + 1$

Step 1

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

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

Step 2

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

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

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 $a = a^{y}$ and $b = 1$ .

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

Step 4

Simplify.

Multiply the exponents in ${(a^{y})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Move $2$ to the left of $y$ .

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

Multiply $- 1$ by $1$ .

$(a^{y} + 1) (a^{2 y} - a^{y} + 1^{2})$

One to any power is one.

$(a^{y} + 1) (a^{2 y} - a^{y} + 1)$

Factor a^(3y)+1

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