Solve 2^3div(-2)+sqrt[3]{5+2-lfloor8div(-2)+2^4rfloor}

$${ 2 }^{ 3 } \div (-2)+ \sqrt[3]{ 5+2- \lfloor 8 \div (-2)+ { 2 }^{ 4 } \rfloor }$$

Evaluate

$$-\sqrt[3]{5}-4\approx -5.709975947$$

Solution Steps

Calculate 2 to the power of 3 and get 8.

$$\frac{8}{-2}+\sqrt[3]{5+2-\lfloor \frac{8}{-2}+2^{4}\rfloor }$$

Divide 8 by -2 to get -4.

$$-4+\sqrt[3]{5+2-\lfloor \frac{8}{-2}+2^{4}\rfloor }$$

Add 5 and 2 to get 7.

$$-4+\sqrt[3]{7-\lfloor \frac{8}{-2}+2^{4}\rfloor }$$

Divide 8 by -2 to get -4.

$$-4+\sqrt[3]{7-\lfloor -4+2^{4}\rfloor }$$

Calculate 2 to the power of 4 and get 16.

$$-4+\sqrt[3]{7-\lfloor -4+16\rfloor }$$

Add -4 and 16 to get 12.

$$-4+\sqrt[3]{7-\lfloor 12\rfloor }$$

The floor of a real number a is the largest integer number less than or equal to a. The floor of 12 is 12.

$$-4+\sqrt[3]{7-12}$$

Subtract 12 from 7 to get -5.

$$-4+\sqrt[3]{-5}$$