Precalculus Examples

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

Step 1

Rewrite $64 x^{6}$ as ${(4 x^{2})}^{3}$ .

${(4 x^{2})}^{3} + 27 y^{3}$

Step 2

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

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

Step 3

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

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

Step 4

Simplify.

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Step 4.1

Apply the product rule to $4 x^{2}$ .

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

Step 4.2

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

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

Step 4.3

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

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Step 4.3.1

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

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

Step 4.3.2

Multiply $2$ by $2$ .

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

Step 4.4

Multiply $4$ by $- 1$ .

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

Step 4.5

Rewrite using the commutative property of multiplication.

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

Step 4.6

Multiply $- 4$ by $3$ .

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

Step 4.7

Apply the product rule to $3 y$ .

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

Step 4.8

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

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