Algebra Examples

x^{6} - 27

$x^{6} - 27$

Step 1

Rewrite

x^{6}

$x^{6}$ as

{(x^{2})}^{3}

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

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

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

Step 2

Rewrite

27

$27$ as

3^{3}

$3^{3}$ .

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

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

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})

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

a = x^{2}

$a = x^{2}$ and

b = 3

$b = 3$ .

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

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

Step 4

Simplify.

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Step 4.1

Multiply the exponents in

{(x^{2})}^{2}

${(x^{2})}^{2}$ .

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Step 4.1.1

Apply the power rule and multiply exponents,

{(a^{m})}^{n} = a^{m n}

${(a^{m})}^{n} = a^{m n}$ .

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

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

Step 4.1.2

Multiply

2

$2$ by

2

$2$ .

(x^{2} - 3) (x^{4} + x^{2} \cdot 3 + 3^{2})

$(x^{2} - 3) (x^{4} + x^{2} \cdot 3 + 3^{2})$

(x^{2} - 3) (x^{4} + x^{2} \cdot 3 + 3^{2})

$(x^{2} - 3) (x^{4} + x^{2} \cdot 3 + 3^{2})$

Step 4.2

Move

3

$3$ to the left of

x^{2}

$x^{2}$ .

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

$(x^{2} - 3) (x^{4} + 3 x^{2} + 3^{2})$

Step 4.3

Raise

3

$3$ to the power of

2

$2$ .

(x^{2} - 3) (x^{4} + 3 x^{2} + 9)

$(x^{2} - 3) (x^{4} + 3 x^{2} + 9)$

(x^{2} - 3) (x^{4} + 3 x^{2} + 9)

$(x^{2} - 3) (x^{4} + 3 x^{2} + 9)$

x^{6} - 27

$x^{6} - 27$

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