Factor 27x^6+8y^3

$27 x^{6} + 8 y^{3}$

Step 1

Rewrite $27 x^{6}$ as ${(3 x^{2})}^{3}$ .

${(3 x^{2})}^{3} + 8 y^{3}$

Step 2

Rewrite $8 y^{3}$ as ${(2 y)}^{3}$ .

${(3 x^{2})}^{3} + {(2 y)}^{3}$

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 $a = 3 x^{2}$ and $b = 2 y$ .

$(3 x^{2} + 2 y) ({(3 x^{2})}^{2} - (3 x^{2}) (2 y) + {(2 y)}^{2})$

Step 4

Simplify.

Apply the product rule to $3 x^{2}$ .

$(3 x^{2} + 2 y) (3^{2} {(x^{2})}^{2} - (3 x^{2}) (2 y) + {(2 y)}^{2})$

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

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

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

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

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

Multiply $2$ by $2$ .

$(3 x^{2} + 2 y) (9 x^{4} - (3 x^{2}) (2 y) + {(2 y)}^{2})$

Multiply $3$ by $- 1$ .

$(3 x^{2} + 2 y) (9 x^{4} - 3 x^{2} (2 y) + {(2 y)}^{2})$

Rewrite using the commutative property of multiplication.

$(3 x^{2} + 2 y) (9 x^{4} - 3 \cdot 2 x^{2} y + {(2 y)}^{2})$

Multiply $- 3$ by $2$ .

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

Apply the product rule to $2 y$ .

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

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

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

Factor 27x^6+8y^3

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