Algebra Examples

64 x^{3} - 27 y^{3}

$64 x^{3} - 27 y^{3}$

Step 1

Rewrite

64 x^{3}

$64 x^{3}$ as

{(4 x)}^{3}

${(4 x)}^{3}$ .

{(4 x)}^{3} - 27 y^{3}

${(4 x)}^{3} - 27 y^{3}$

Step 2

Rewrite

27 y^{3}

$27 y^{3}$ as

{(3 y)}^{3}

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

{(4 x)}^{3} - {(3 y)}^{3}

${(4 x)}^{3} - {(3 y)}^{3}$

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 = 4 x

$a = 4 x$ and

b = 3 y

$b = 3 y$ .

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

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

Step 4

Simplify.

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Step 4.1

Multiply

3

$3$ by

- 1

$- 1$ .

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

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

Step 4.2

Apply the product rule to

4 x

$4 x$ .

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

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

Step 4.3

Raise

4

$4$ to the power of

2

$2$ .

(4 x - 3 y) (16 x^{2} + 4 x (3 y) + {(3 y)}^{2})

$(4 x - 3 y) (16 x^{2} + 4 x (3 y) + {(3 y)}^{2})$

Step 4.4

Rewrite using the commutative property of multiplication.

(4 x - 3 y) (16 x^{2} + 4 \cdot 3 x y + {(3 y)}^{2})

$(4 x - 3 y) (16 x^{2} + 4 \cdot 3 x y + {(3 y)}^{2})$

Step 4.5

Multiply

4

$4$ by

3

$3$ .

(4 x - 3 y) (16 x^{2} + 12 x y + {(3 y)}^{2})

$(4 x - 3 y) (16 x^{2} + 12 x y + {(3 y)}^{2})$

Step 4.6

Apply the product rule to

3 y

$3 y$ .

(4 x - 3 y) (16 x^{2} + 12 x y + 3^{2} y^{2})

$(4 x - 3 y) (16 x^{2} + 12 x y + 3^{2} y^{2})$

Step 4.7

Raise

3

$3$ to the power of

2

$2$ .

(4 x - 3 y) (16 x^{2} + 12 x y + 9 y^{2})

$(4 x - 3 y) (16 x^{2} + 12 x y + 9 y^{2})$

(4 x - 3 y) (16 x^{2} + 12 x y + 9 y^{2})

$(4 x - 3 y) (16 x^{2} + 12 x y + 9 y^{2})$