Factor x^6+2x^3-3

$x^{6} + 2 x^{3} - 3$

Step 1

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

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

Step 2

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

$u^{2} + 2 u - 3$

Step 3

Factor $u^{2} + 2 u - 3$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 3$ and whose sum is $2$ .

$- 1, 3$

Write the factored form using these integers.

$(u - 1) (u + 3)$

Step 4

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

$(x^{3} - 1) (x^{3} + 3)$

Step 5

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

$(x^{3} - 1^{3}) (x^{3} + 3)$

Step 6

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 $a = x$ and $b = 1$ .

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

Step 7

Simplify.

Multiply $x$ by $1$ .

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

One to any power is one.

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

Factor x^6+2x^3-3

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