Algebra Examples

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

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

Step 1

Rewrite

z^{6}

$z^{6}$ as

{(z^{3})}^{2}

${(z^{3})}^{2}$ .

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

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

Step 2

Let

u = z^{3}

$u = z^{3}$ . Substitute

u

$u$ for all occurrences of

z^{3}

$z^{3}$ .

u^{2} - 9 u + 8

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

Step 3

Factor

u^{2} - 9 u + 8

$u^{2} - 9 u + 8$ using the AC method.

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Step 3.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

8

$8$ and whose sum is

- 9

$- 9$ .

- 8, - 1

$- 8, - 1$

Step 3.2

Write the factored form using these integers.

(u - 8) (u - 1)

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

(u - 8) (u - 1)

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

Step 4

Replace all occurrences of

u

$u$ with

z^{3}

$z^{3}$ .

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

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

Step 5

Rewrite

8

$8$ as

2^{3}

$2^{3}$ .

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

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

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})

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

a = z

$a = z$ and

b = 2

$b = 2$ .

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

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

Step 7

Simplify.

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Step 7.1

Move

2

$2$ to the left of

z

$z$ .

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

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

Step 7.2

Raise

2

$2$ to the power of

2

$2$ .

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

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

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

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

Step 8

Rewrite

1

$1$ as

1^{3}

$1^{3}$ .

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

$a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where

a = z

$a = z$ and

b = 1

$b = 1$ .

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

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

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.

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