Algebra Examples

27 a^{3} - 125 b^{3}

$27 a^{3} - 125 b^{3}$

Step 1

Rewrite

27 a^{3}

$27 a^{3}$ as

{(3 a)}^{3}

${(3 a)}^{3}$ .

{(3 a)}^{3} - 125 b^{3}

${(3 a)}^{3} - 125 b^{3}$

Step 2

Rewrite

125 b^{3}

$125 b^{3}$ as

{(5 b)}^{3}

${(5 b)}^{3}$ .

{(3 a)}^{3} - {(5 b)}^{3}

${(3 a)}^{3} - {(5 b)}^{3}$

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 = 3 a

$a = 3 a$ and

b = 5 b

$b = 5 b$ .

(3 a - (5 b)) ({(3 a)}^{2} + 3 a (5 b) + {(5 b)}^{2})

$(3 a - (5 b)) ({(3 a)}^{2} + 3 a (5 b) + {(5 b)}^{2})$

Step 4

Simplify.

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Step 4.1

Multiply

5

$5$ by

- 1

$- 1$ .

(3 a - 5 b) ({(3 a)}^{2} + 3 a (5 b) + {(5 b)}^{2})

$(3 a - 5 b) ({(3 a)}^{2} + 3 a (5 b) + {(5 b)}^{2})$

Step 4.2

Apply the product rule to

3 a

$3 a$ .

(3 a - 5 b) (3^{2} a^{2} + 3 a (5 b) + {(5 b)}^{2})

$(3 a - 5 b) (3^{2} a^{2} + 3 a (5 b) + {(5 b)}^{2})$

Step 4.3

Raise

3

$3$ to the power of

2

$2$ .

(3 a - 5 b) (9 a^{2} + 3 a (5 b) + {(5 b)}^{2})

$(3 a - 5 b) (9 a^{2} + 3 a (5 b) + {(5 b)}^{2})$

Step 4.4

Rewrite using the commutative property of multiplication.

(3 a - 5 b) (9 a^{2} + 3 \cdot 5 a b + {(5 b)}^{2})

$(3 a - 5 b) (9 a^{2} + 3 \cdot 5 a b + {(5 b)}^{2})$

Step 4.5

Multiply

3

$3$ by

5

$5$ .

(3 a - 5 b) (9 a^{2} + 15 a b + {(5 b)}^{2})

$(3 a - 5 b) (9 a^{2} + 15 a b + {(5 b)}^{2})$

Step 4.6

Apply the product rule to

5 b

$5 b$ .

(3 a - 5 b) (9 a^{2} + 15 a b + 5^{2} b^{2})

$(3 a - 5 b) (9 a^{2} + 15 a b + 5^{2} b^{2})$

Step 4.7

Raise

5

$5$ to the power of

2

$2$ .

(3 a - 5 b) (9 a^{2} + 15 a b + 25 b^{2})

$(3 a - 5 b) (9 a^{2} + 15 a b + 25 b^{2})$

(3 a - 5 b) (9 a^{2} + 15 a b + 25 b^{2})

$(3 a - 5 b) (9 a^{2} + 15 a b + 25 b^{2})$