Factor x^6+16x^3+64

$x^{6} + 16 x^{3} + 64$

Step 1

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

${(x^{3})}^{2} + 16 x^{3} + 64$

Step 2

Let $u = x^{3}$ . Substitute $u$ for all occurrences of $x^{3}$ .

$u^{2} + 16 u + 64$

Step 3

Factor using the perfect square rule.

Rewrite $64$ as $8^{2}$ .

$u^{2} + 16 u + 8^{2}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$16 u = 2 \cdot u \cdot 8$

Rewrite the polynomial.

$u^{2} + 2 \cdot u \cdot 8 + 8^{2}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 8$ .

${(u + 8)}^{2}$

Step 4

Replace all occurrences of $u$ with $x^{3}$ .

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

Step 5

Rewrite $8$ as $2^{3}$ .

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

Step 6

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 $a = x$ and $b = 2$ .

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

Step 7

Simplify.

Multiply $2$ by $- 1$ .

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

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

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

Step 8

Apply the product rule to $(x + 2) (x^{2} - 2 x + 4)$ .

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

Factor x^6+16x^3+64

Related Questions

Support LCMGCF.com by disabling your adblocker.