Algebra Examples

$z^{6} - 9 z^{3} + 8$

Step 1

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

${(z^{3})}^{2} - 9 z^{3} + 8$

Step 2

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

$u^{2} - 9 u + 8$

Step 3

Factor $u^{2} - 9 u + 8$ 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 $8$ and whose sum is $- 9$ .

$- 8, - 1$

Step 3.2

Write the factored form using these integers.

$(u - 8) (u - 1)$

Step 4

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

$(z^{3} - 8) (z^{3} - 1)$

Step 5

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

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

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 $a = z$ and $b = 2$ .

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

Step 7

Simplify.

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Step 7.1

Move $2$ to the left of $z$ .

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

Step 7.2

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

$(z - 2) (z^{2} + 2 z + 4) (z^{3} - 1)$

Step 8

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

$(z - 2) (z^{2} + 2 z + 4) (z^{3} - 1^{3})$

Step 9

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 $a = z$ and $b = 1$ .

$(z - 2) (z^{2} + 2 z + 4) ((z - 1) (z^{2} + z \cdot 1 + 1^{2}))$

Step 10

Factor.

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Step 10.1

Simplify.

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Step 10.1.1

Multiply $z$ by $1$ .

$(z - 2) (z^{2} + 2 z + 4) ((z - 1) (z^{2} + z + 1^{2}))$

Step 10.1.2

One to any power is one.

$(z - 2) (z^{2} + 2 z + 4) ((z - 1) (z^{2} + z + 1))$

Step 10.2

Remove unnecessary parentheses.

$(z - 2) (z^{2} + 2 z + 4) (z - 1) (z^{2} + z + 1)$