Basic Math Examples

$192 y^{3} + 81 z^{3}$

Step 1

Factor $3$ out of $192 y^{3} + 81 z^{3}$ .

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Step 1.1

Factor $3$ out of $192 y^{3}$ .

$3 (64 y^{3}) + 81 z^{3}$

Step 1.2

Factor $3$ out of $81 z^{3}$ .

$3 (64 y^{3}) + 3 (27 z^{3})$

Step 1.3

Factor $3$ out of $3 (64 y^{3}) + 3 (27 z^{3})$ .

$3 (64 y^{3} + 27 z^{3})$

Step 2

Rewrite $64 y^{3}$ as ${(4 y)}^{3}$ .

$3 ({(4 y)}^{3} + 27 z^{3})$

Step 3

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

$3 ({(4 y)}^{3} + {(3 z)}^{3})$

Step 4

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 $a = 4 y$ and $b = 3 z$ .

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

Step 5

Factor.

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Step 5.1

Simplify.

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Step 5.1.1

Apply the product rule to $4 y$ .

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

Step 5.1.2

Raise $4$ to the power of $2$ .

$3 ((4 y + 3 z) (16 y^{2} - (4 y) (3 z) + {(3 z)}^{2}))$

Step 5.1.3

Multiply $4$ by $- 1$ .

$3 ((4 y + 3 z) (16 y^{2} - 4 y (3 z) + {(3 z)}^{2}))$

Step 5.1.4

Rewrite using the commutative property of multiplication.

$3 ((4 y + 3 z) (16 y^{2} - 4 \cdot 3 y z + {(3 z)}^{2}))$

Step 5.1.5

Multiply $- 4$ by $3$ .

$3 ((4 y + 3 z) (16 y^{2} - 12 y z + {(3 z)}^{2}))$

Step 5.1.6

Apply the product rule to $3 z$ .

$3 ((4 y + 3 z) (16 y^{2} - 12 y z + 3^{2} z^{2}))$

Step 5.1.7

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

$3 ((4 y + 3 z) (16 y^{2} - 12 y z + 9 z^{2}))$

Step 5.2

Remove unnecessary parentheses.

$3 (4 y + 3 z) (16 y^{2} - 12 y z + 9 z^{2})$