Solve -2^4-sqrt{36}+6div(-2/3)*sqrt[3]{-8}

$$- 2 ^ { 4 } - \sqrt { 36 } + 6 \div ( - \frac { 2 } { 3 } ) \times \sqrt[ 3 ] { - 8 }$$

Evaluate

$$-4$$

Solution Steps

Calculate 2 to the power of 4 and get 16.

$$-16-\sqrt{36}+\frac{6}{-\frac{2}{3}}\sqrt[3]{-8}$$

Calculate the square root of 36 and get 6.

$$-16-6+\frac{6}{-\frac{2}{3}}\sqrt[3]{-8}$$

Subtract 6 from -16 to get -22.

$$-22+\frac{6}{-\frac{2}{3}}\sqrt[3]{-8}$$

Divide 6 by -\frac{2}{3} by multiplying 6 by the reciprocal of -\frac{2}{3}.

$$-22+6\left(-\frac{3}{2}\right)\sqrt[3]{-8}$$

Multiply 6 and -\frac{3}{2} to get -9.

$$-22-9\sqrt[3]{-8}$$

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

$$-22-9\left(-2\right)$$

Multiply -9 and -2 to get 18.

$$-22+18$$

Add -22 and 18 to get -4.

$$-4$$

Factor

$$-4$$