Solve sqrt[3]{646/125-11}-sqrt{0.1^-2}+sqrt[3]{8^2}

$$\sqrt[ 3 ] { \frac { 646 } { 125 } - 11 } - \sqrt { 0.1 ^ { - 2 } } + \sqrt[ 3 ] { 8 ^ { 2 } }$$

Evaluate

$$-7.8$$

Solution Steps

Subtract 11 from \frac{646}{125} to get -\frac{729}{125}.

$$\sqrt[3]{-\frac{729}{125}}-\sqrt{0.1^{-2}}+\sqrt[3]{8^{2}}$$

Calculate \sqrt[3]{-\frac{729}{125}} and get -\frac{9}{5}.

$$-\frac{9}{5}-\sqrt{0.1^{-2}}+\sqrt[3]{8^{2}}$$

Calculate 0.1 to the power of -2 and get 100.

$$-\frac{9}{5}-\sqrt{100}+\sqrt[3]{8^{2}}$$

Calculate the square root of 100 and get 10.

$$-\frac{9}{5}-10+\sqrt[3]{8^{2}}$$

Subtract 10 from -\frac{9}{5} to get -\frac{59}{5}.

$$-\frac{59}{5}+\sqrt[3]{8^{2}}$$

Calculate 8 to the power of 2 and get 64.

$$-\frac{59}{5}+\sqrt[3]{64}$$

Calculate \sqrt[3]{64} and get 4.

$$-\frac{59}{5}+4$$

Add -\frac{59}{5} and 4 to get -\frac{39}{5}.

$$-\frac{39}{5}$$

Factor

$$-7.8$$