Basic Math Examples

$27 v^{6} + 64 g^{3}$

Step 1

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

${(3 v^{2})}^{3} + 64 g^{3}$

Step 2

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

${(3 v^{2})}^{3} + {(4 g)}^{3}$

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 $a = 3 v^{2}$ and $b = 4 g$ .

$(3 v^{2} + 4 g) ({(3 v^{2})}^{2} - (3 v^{2}) (4 g) + {(4 g)}^{2})$

Step 4

Simplify.

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Step 4.1

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

$(3 v^{2} + 4 g) (3^{2} {(v^{2})}^{2} - (3 v^{2}) (4 g) + {(4 g)}^{2})$

Step 4.2

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

$(3 v^{2} + 4 g) (9 {(v^{2})}^{2} - (3 v^{2}) (4 g) + {(4 g)}^{2})$

Step 4.3

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

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Step 4.3.1

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

$(3 v^{2} + 4 g) (9 v^{2 \cdot 2} - (3 v^{2}) (4 g) + {(4 g)}^{2})$

Step 4.3.2

Multiply $2$ by $2$ .

$(3 v^{2} + 4 g) (9 v^{4} - (3 v^{2}) (4 g) + {(4 g)}^{2})$

Step 4.4

Multiply $3$ by $- 1$ .

$(3 v^{2} + 4 g) (9 v^{4} - 3 v^{2} (4 g) + {(4 g)}^{2})$

Step 4.5

Rewrite using the commutative property of multiplication.

$(3 v^{2} + 4 g) (9 v^{4} - 3 \cdot 4 v^{2} g + {(4 g)}^{2})$

Step 4.6

Multiply $- 3$ by $4$ .

$(3 v^{2} + 4 g) (9 v^{4} - 12 v^{2} g + {(4 g)}^{2})$

Step 4.7

Apply the product rule to $4 g$ .

$(3 v^{2} + 4 g) (9 v^{4} - 12 v^{2} g + 4^{2} g^{2})$

Step 4.8

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

$(3 v^{2} + 4 g) (9 v^{4} - 12 v^{2} g + 16 g^{2})$