Factor x^6+27

$x^{6} + 27$

Step 1

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

${(x^{2})}^{3} + 27$

Step 2

Rewrite $27$ as $3^{3}$ .

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

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

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

Step 4

Simplify.

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

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

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

Multiply $2$ by $2$ .

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

Multiply $3$ by $- 1$ .

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

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

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

Factor.

Rewrite $x^{4} - 3 x^{2} + 9$ in a factored form.

Rewrite the middle term.

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

Rearrange terms.

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

Factor first three terms by perfect square rule.

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

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

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

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

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

Simplify.

Reorder terms.

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

Multiply $3$ by $- 1$ .

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

Reorder terms.

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

Remove unnecessary parentheses.

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

Factor x^6+27

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