Solve [sqrt[3]{sqrt{64}}-3]*(1-2^2)

$$[ \sqrt[ 3 ] { \sqrt { 64 } } - 3 ] \cdot ( 1 - 2 ^ { 2 } )$$

Evaluate

$$3$$

Solution Steps

Calculate the square root of 64 and get 8.

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

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

$$\left(2-3\right)\left(1-2^{2}\right)$$

Subtract 3 from 2 to get -1.

$$-\left(1-2^{2}\right)$$

Calculate 2 to the power of 2 and get 4.

$$-\left(1-4\right)$$

Subtract 4 from 1 to get -3.

$$-\left(-3\right)$$

Multiply -1 and -3 to get 3.

$$3$$

Factor

$$3$$