Finite Math Examples

$27 x^{6} - 64 y^{3}$

Step 1

Rewrite $27 x^{6}$ as ${(3 x^{2})}^{3}$ .

${(3 x^{2})}^{3} - 64 y^{3}$

Step 2

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

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

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 $a = 3 x^{2}$ and $b = 4 y$ .

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

Step 4

Simplify.

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Step 4.1

Multiply $4$ by $- 1$ .

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

Step 4.2

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

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

Step 4.3

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

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

Step 4.4

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

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Step 4.4.1

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

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

Step 4.4.2

Multiply $2$ by $2$ .

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

Step 4.5

Rewrite using the commutative property of multiplication.

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

Step 4.6

Multiply $3$ by $4$ .

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

Step 4.7

Apply the product rule to $4 y$ .

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

Step 4.8

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

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