Precalculus Examples

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

Step 1

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

${(2 x)}^{3} + y^{6}$

Step 2

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

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

Step 3

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 $a = 2 x$ and $b = y^{2}$ .

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

Step 4

Simplify.

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Step 4.1

Apply the product rule to $2 x$ .

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

Step 4.2

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

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

Step 4.3

Multiply $2$ by $- 1$ .

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

Step 4.4

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

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Step 4.4.1

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

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

Step 4.4.2

Multiply $2$ by $2$ .

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