Algebra Examples

$x^{3} + 8 y^{6}$

Step 1

Rewrite $8 y^{6}$ as ${(2 y^{2})}^{3}$ .

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

Step 2

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = x$ and $b = 2 y^{2}$ .

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

Step 3

Simplify.

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Step 3.1

Rewrite using the commutative property of multiplication.

$(x + 2 y^{2}) (x^{2} - 1 \cdot 2 x y^{2} + {(2 y^{2})}^{2})$

Step 3.2

Multiply $- 1$ by $2$ .

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

Step 3.3

Apply the product rule to $2 y^{2}$ .

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

Step 3.4

Raise $2$ to the power of $2$ .

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

Step 3.5

Multiply the exponents in ${(y^{2})}^{2}$ .

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Step 3.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 3.5.2

Multiply $2$ by $2$ .

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