Algebra Examples

x^{3} - 8 y^{3}

$x^{3} - 8 y^{3}$

Step 1

Rewrite

8 y^{3}

$8 y^{3}$ as

{(2 y)}^{3}

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

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

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

Step 2

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

$a = x$ and

b = 2 y

$b = 2 y$ .

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

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

Step 3

Simplify.

Tap for more steps...

Step 3.1

Multiply

2

$2$ by

- 1

$- 1$ .

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

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

Step 3.2

Rewrite using the commutative property of multiplication.

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

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

Step 3.3

Apply the product rule to

2 y

$2 y$ .

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

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

Step 3.4

Raise

2

$2$ to the power of

2

$2$ .

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

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

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

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

x^{3} - 8 y^{3}

$x^{3} - 8 y^{3}$

(

)

[

]

√



≥

















≤















∩

∪







∞











