Solve frac{sqrt{9}-sqrt[3]{-8}+(1/2)^0}{(-2)^2+sqrt[3]{-27}}=

$$\frac { \sqrt { 9 } - \sqrt[ 3 ] { - 8 } + ( \frac { 1 } { 2 } ) ^ { 0 } } { ( - 2 ) ^ { 2 } + \sqrt[ 3 ] { - 27 } } =$$

Evaluate

$$6$$

Solution Steps

Calculate the square root of 9 and get 3.

$$\frac{3-\sqrt[3]{-8}+\left(\frac{1}{2}\right)^{0}}{\left(-2\right)^{2}+\sqrt[3]{-27}}$$

Calculate \sqrt[3]{-8} and get -2.

$$\frac{3-\left(-2\right)+\left(\frac{1}{2}\right)^{0}}{\left(-2\right)^{2}+\sqrt[3]{-27}}$$

The opposite of -2 is 2.

$$\frac{3+2+\left(\frac{1}{2}\right)^{0}}{\left(-2\right)^{2}+\sqrt[3]{-27}}$$

Add 3 and 2 to get 5.

$$\frac{5+\left(\frac{1}{2}\right)^{0}}{\left(-2\right)^{2}+\sqrt[3]{-27}}$$

Calculate \frac{1}{2} to the power of 0 and get 1.

$$\frac{5+1}{\left(-2\right)^{2}+\sqrt[3]{-27}}$$

Add 5 and 1 to get 6.

$$\frac{6}{\left(-2\right)^{2}+\sqrt[3]{-27}}$$

Calculate -2 to the power of 2 and get 4.

$$\frac{6}{4+\sqrt[3]{-27}}$$

Calculate \sqrt[3]{-27} and get -3.

$$\frac{6}{4-3}$$

Subtract 3 from 4 to get 1.

$$\frac{6}{1}$$

Anything divided by one gives itself.

$$6$$

Factor

$$2\times 3$$