Algebra Examples

$x^{6} - 26 x^{3} - 27$

Step 1

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

${(x^{3})}^{2} - 26 x^{3} - 27$

Step 2

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

$u^{2} - 26 u - 27$

Step 3

Factor $u^{2} - 26 u - 27$ using the AC method.

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Step 3.1

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 $- 27$ and whose sum is $- 26$ .

$- 27, 1$

Step 3.2

Write the factored form using these integers.

$(u - 27) (u + 1)$

Step 4

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

$(x^{3} - 27) (x^{3} + 1)$

Step 5

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

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

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 $b = 3$ .

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

Step 7

Simplify.

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Step 7.1

Move $3$ to the left of $x$ .

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

Step 7.2

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

$(x - 3) (x^{2} + 3 x + 9) (x^{3} + 1)$

Step 8

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

$(x - 3) (x^{2} + 3 x + 9) (x^{3} + 1^{3})$

Step 9

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 $b = 1$ .

$(x - 3) (x^{2} + 3 x + 9) ((x + 1) (x^{2} - x \cdot 1 + 1^{2}))$

Step 10

Factor.

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Step 10.1

Simplify.

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Step 10.1.1

Multiply $- 1$ by $1$ .

$(x - 3) (x^{2} + 3 x + 9) ((x + 1) (x^{2} - x + 1^{2}))$

Step 10.1.2

One to any power is one.

$(x - 3) (x^{2} + 3 x + 9) ((x + 1) (x^{2} - x + 1))$

Step 10.2

Remove unnecessary parentheses.

$(x - 3) (x^{2} + 3 x + 9) (x + 1) (x^{2} - x + 1)$