Solve 4-sqrt{16}+[frac{2}2^3*(7*2-4^2)-3]-2*sqrt[3]{8}

$$4 - \sqrt { 16 } + [ \frac { 2 } { 2 } ^ { 3 } \cdot ( 7 \cdot 2 - 4 ^ { 2 } ) - 3 ] - 2 \cdot \sqrt[ 3 ] { 8 }$$

Evaluate

$$-9$$

Solution Steps

Divide 2 by 2 to get 1.

$$4-\sqrt{16}+1^{3}\left(7\times 2-4^{2}\right)-3-2\sqrt[3]{8}$$

Calculate the square root of 16 and get 4.

$$4-4+1^{3}\left(7\times 2-4^{2}\right)-3-2\sqrt[3]{8}$$

Subtract 4 from 4 to get 0.

$$1^{3}\left(7\times 2-4^{2}\right)-3-2\sqrt[3]{8}$$

Calculate 1 to the power of 3 and get 1.

$$1\left(7\times 2-4^{2}\right)-3-2\sqrt[3]{8}$$

Multiply 7 and 2 to get 14.

$$1\left(14-4^{2}\right)-3-2\sqrt[3]{8}$$

Calculate 4 to the power of 2 and get 16.

$$1\left(14-16\right)-3-2\sqrt[3]{8}$$

Subtract 16 from 14 to get -2.

$$1\left(-2\right)-3-2\sqrt[3]{8}$$

Multiply 1 and -2 to get -2.

$$-2-3-2\sqrt[3]{8}$$

Subtract 3 from -2 to get -5.

$$-5-2\sqrt[3]{8}$$

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

$$-5-2\times 2$$

Multiply -2 and 2 to get -4.

$$-5-4$$

Subtract 4 from -5 to get -9.

$$-9$$

Factor

$$-9$$