Basic Math Examples

$\frac{1}{6} \cdot (\sqrt{36} \div 3^{- 2}) + 4^{3} \div | - 8 |$

Step 1

Simplify each term.

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Step 1.1

Combine.

$\frac{1 \sqrt{36}}{6 \cdot 3^{- 2}} + 4^{3} \div | - 8 |$

Step 1.2

Move $3^{- 2}$ to the numerator using the negative exponent rule $\frac{1}{b^{- n}} = b^{n}$ .

$\frac{\sqrt{36} \cdot 3^{2}}{6} + 4^{3} \div | - 8 |$

Step 1.3

Simplify the numerator.

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Step 1.3.1

Rewrite $36$ as $6^{2}$ .

$\frac{\sqrt{6^{2}} \cdot 3^{2}}{6} + 4^{3} \div | - 8 |$

Step 1.3.2

Pull terms out from under the radical, assuming positive real numbers.

$\frac{6 \cdot 3^{2}}{6} + 4^{3} \div | - 8 |$

Step 1.3.3

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

$\frac{6 \cdot 9}{6} + 4^{3} \div | - 8 |$

Step 1.4

Multiply $6$ by $9$ .

$\frac{54}{6} + 4^{3} \div | - 8 |$

Step 1.5

Divide $54$ by $6$ .

$9 + 4^{3} \div | - 8 |$

Step 1.6

Rewrite the division as a fraction.

$9 + \frac{4^{3}}{| - 8 |}$

Step 1.7

Raise $4$ to the power of $3$ .

$9 + \frac{64}{| - 8 |}$

Step 1.8

The absolute value is the distance between a number and zero. The distance between $- 8$ and $0$ is $8$ .

$9 + \frac{64}{8}$

Step 1.9

Divide $64$ by $8$ .

$9 + 8$

Step 2

Add $9$ and $8$ .

$17$