Algebra Examples

$8 x^{5} y^{2} - 27 x^{2} y^{5}$

Step 1

Factor $x^{2} y^{2}$ out of $8 x^{5} y^{2} - 27 x^{2} y^{5}$ .

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Step 1.1

Factor $x^{2} y^{2}$ out of $8 x^{5} y^{2}$ .

$x^{2} y^{2} (8 x^{3}) - 27 x^{2} y^{5}$

Step 1.2

Factor $x^{2} y^{2}$ out of $- 27 x^{2} y^{5}$ .

$x^{2} y^{2} (8 x^{3}) + x^{2} y^{2} (- 27 y^{3})$

Step 1.3

Factor $x^{2} y^{2}$ out of $x^{2} y^{2} (8 x^{3}) + x^{2} y^{2} (- 27 y^{3})$ .

$x^{2} y^{2} (8 x^{3} - 27 y^{3})$

Step 2

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

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

Step 3

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

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

Step 4

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 $a = 2 x$ and $b = 3 y$ .

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

Step 5

Factor.

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Step 5.1

Simplify.

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Step 5.1.1

Multiply $3$ by $- 1$ .

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

Step 5.1.2

Apply the product rule to $2 x$ .

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

Step 5.1.3

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

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

Step 5.1.4

Rewrite using the commutative property of multiplication.

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

Step 5.1.5

Multiply $2$ by $3$ .

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

Step 5.1.6

Apply the product rule to $3 y$ .

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

Step 5.1.7

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

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

Step 5.2

Remove unnecessary parentheses.

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