Factor 27/125x^6-8y^9

$\frac{27}{125} x^{6} - 8 y^{9}$

Step 1

Rewrite $\frac{27}{125} x^{6}$ as ${(\frac{3}{5} x^{2})}^{3}$ .

${(\frac{3}{5} x^{2})}^{3} - 8 y^{9}$

Step 2

Rewrite $8 y^{9}$ as ${(2 y^{3})}^{3}$ .

${(\frac{3}{5} x^{2})}^{3} - {(2 y^{3})}^{3}$

Step 3

Since both terms are perfect cubes, factor using the difference of cubes formula, $a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})$ where $a = \frac{3}{5} x^{2}$ and $b = 2 y^{3}$ .

$(\frac{3}{5} x^{2} - (2 y^{3})) ({(\frac{3}{5} x^{2})}^{2} + \frac{3}{5} x^{2} (2 y^{3}) + {(2 y^{3})}^{2})$

Step 4

Simplify.

Combine $\frac{3}{5}$ and $x^{2}$ .

$(\frac{3 x^{2}}{5} - (2 y^{3})) ({(\frac{3}{5} x^{2})}^{2} + \frac{3}{5} x^{2} (2 y^{3}) + {(2 y^{3})}^{2})$

Multiply $2$ by $- 1$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) ({(\frac{3}{5} x^{2})}^{2} + \frac{3}{5} x^{2} (2 y^{3}) + {(2 y^{3})}^{2})$

Combine $\frac{3}{5}$ and $x^{2}$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) ({(\frac{3 x^{2}}{5})}^{2} + \frac{3}{5} x^{2} (2 y^{3}) + {(2 y^{3})}^{2})$

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

Apply the product rule to $\frac{3 x^{2}}{5}$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{{(3 x^{2})}^{2}}{5^{2}} + \frac{3}{5} x^{2} (2 y^{3}) + {(2 y^{3})}^{2})$

Apply the product rule to $3 x^{2}$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{3^{2} {(x^{2})}^{2}}{5^{2}} + \frac{3}{5} x^{2} (2 y^{3}) + {(2 y^{3})}^{2})$

Simplify the numerator.

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

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 {(x^{2})}^{2}}{5^{2}} + \frac{3}{5} x^{2} (2 y^{3}) + {(2 y^{3})}^{2})$

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

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

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{2 \cdot 2}}{5^{2}} + \frac{3}{5} x^{2} (2 y^{3}) + {(2 y^{3})}^{2})$

Multiply $2$ by $2$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{5^{2}} + \frac{3}{5} x^{2} (2 y^{3}) + {(2 y^{3})}^{2})$

Raise $5$ to the power of $2$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{25} + \frac{3}{5} x^{2} (2 y^{3}) + {(2 y^{3})}^{2})$

Rewrite using the commutative property of multiplication.

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{25} + \frac{3}{5} \cdot 2 x^{2} y^{3} + {(2 y^{3})}^{2})$

Multiply $\frac{3}{5} \cdot 2$ .

Combine $\frac{3}{5}$ and $2$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{25} + \frac{3 \cdot 2}{5} x^{2} y^{3} + {(2 y^{3})}^{2})$

Multiply $3$ by $2$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{25} + \frac{6}{5} x^{2} y^{3} + {(2 y^{3})}^{2})$

Combine $\frac{6}{5}$ and $x^{2}$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{25} + \frac{6 x^{2}}{5} y^{3} + {(2 y^{3})}^{2})$

Combine $\frac{6 x^{2}}{5}$ and $y^{3}$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{25} + \frac{6 x^{2} y^{3}}{5} + {(2 y^{3})}^{2})$

Apply the product rule to $2 y^{3}$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{25} + \frac{6 x^{2} y^{3}}{5} + 2^{2} {(y^{3})}^{2})$

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

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{25} + \frac{6 x^{2} y^{3}}{5} + 4 {(y^{3})}^{2})$

Multiply the exponents in ${(y^{3})}^{2}$ .

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

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{25} + \frac{6 x^{2} y^{3}}{5} + 4 y^{3 \cdot 2})$

Multiply $3$ by $2$ .

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{9 x^{4}}{25} + \frac{6 x^{2} y^{3}}{5} + 4 y^{6})$

Factor.

Factor $\frac{1}{25}$ out of each term.

$(\frac{3 x^{2}}{5} - 2 y^{3}) (\frac{1}{25} (9 x^{4} + 30 x^{2} y^{3} + 100 y^{6}))$

Remove unnecessary parentheses.

$(\frac{3 x^{2}}{5} - 2 y^{3}) \frac{1}{25} (9 x^{4} + 30 x^{2} y^{3} + 100 y^{6})$

Factor 27/125x^6-8y^9

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