Factor over the Complex Numbers 1/8x^3-1/27y^3

$\frac{1}{8} x^{3} - \frac{1}{27} y^{3}$

Step 1

Rewrite $\frac{1}{8} x^{3}$ as ${(\frac{1}{2} x)}^{3}$ .

${(\frac{1}{2} x)}^{3} - \frac{1}{27} y^{3}$

Step 2

Rewrite $\frac{1}{27} y^{3}$ as ${(\frac{1}{3} y)}^{3}$ .

${(\frac{1}{2} x)}^{3} - {(\frac{1}{3} y)}^{3}$

Step 3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - i^{3} = (a - i) (a^{2} + a i + i^{2})$ where $a = \frac{1}{2} x$ and $i = \frac{1}{3} y$ .

$(\frac{1}{2} x - (\frac{1}{3} y)) ({(\frac{1}{2} x)}^{2} + \frac{1}{2} x (\frac{1}{3} y) + {(\frac{1}{3} y)}^{2})$

Step 4

Simplify.

Combine $\frac{1}{2}$ and $x$ .

$(\frac{x}{2} - (\frac{1}{3} y)) ({(\frac{1}{2} x)}^{2} + \frac{1}{2} x (\frac{1}{3} y) + {(\frac{1}{3} y)}^{2})$

Combine $\frac{1}{3}$ and $y$ .

$(\frac{x}{2} - \frac{y}{3}) ({(\frac{1}{2} x)}^{2} + \frac{1}{2} x (\frac{1}{3} y) + {(\frac{1}{3} y)}^{2})$

Combine $\frac{1}{2}$ and $x$ .

$(\frac{x}{2} - \frac{y}{3}) ({(\frac{x}{2})}^{2} + \frac{1}{2} x (\frac{1}{3} y) + {(\frac{1}{3} y)}^{2})$

Apply the product rule to $\frac{x}{2}$ .

$(\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{2^{2}} + \frac{1}{2} x (\frac{1}{3} y) + {(\frac{1}{3} y)}^{2})$

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

$(\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{4} + \frac{1}{2} x (\frac{1}{3} y) + {(\frac{1}{3} y)}^{2})$

Rewrite using the commutative property of multiplication.

$(\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{4} + \frac{1}{2} \cdot \frac{1}{3} x y + {(\frac{1}{3} y)}^{2})$

Multiply $\frac{1}{2} \cdot \frac{1}{3}$ .

Multiply $\frac{1}{2}$ by $\frac{1}{3}$ .

$(\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{4} + \frac{1}{2 \cdot 3} x y + {(\frac{1}{3} y)}^{2})$

Multiply $2$ by $3$ .

$(\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{4} + \frac{1}{6} x y + {(\frac{1}{3} y)}^{2})$

Combine $\frac{1}{6}$ and $x$ .

$(\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{4} + \frac{x}{6} y + {(\frac{1}{3} y)}^{2})$

Combine $\frac{x}{6}$ and $y$ .

$(\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{4} + \frac{x y}{6} + {(\frac{1}{3} y)}^{2})$

Combine $\frac{1}{3}$ and $y$ .

$(\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{4} + \frac{x y}{6} + {(\frac{y}{3})}^{2})$

Apply the product rule to $\frac{y}{3}$ .

$(\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{4} + \frac{x y}{6} + \frac{y^{2}}{3^{2}})$

Raise $3$ to the power of $2$ .

$(\frac{x}{2} - \frac{y}{3}) (\frac{x^{2}}{4} + \frac{x y}{6} + \frac{y^{2}}{9})$

Factor over the Complex Numbers 1/8x^3-1/27y^3

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