Factor 1-2a^3+a^6

$1 - 2 a^{3} + a^{6}$

Step 1

Factor using the perfect square rule.

Rewrite $1$ as $1^{2}$ .

$1^{2} - 2 a^{3} + a^{6}$

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

$1^{2} - 2 a^{3} + {(a^{3})}^{2}$

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

$2 a^{3} = 2 \cdot 1 \cdot a^{3}$

Rewrite the polynomial.

$1^{2} - 2 \cdot 1 \cdot a^{3} + {(a^{3})}^{2}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 1$ and $b = a^{3}$ .

${(1 - a^{3})}^{2}$

Step 2

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

${(1^{3} - a^{3})}^{2}$

Step 3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = 1$ and $b = a$ .

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

Step 4

Simplify.

One to any power is one.

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

Multiply $a$ by $1$ .

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

Step 5

Apply the product rule to $(1 - a) (1 + a + a^{2})$ .

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

Factor 1-2a^3+a^6

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