Solve (-2)^6*4^-3-(-2)^3-(1/4)^-1?

$$( - 2 ) ^ { 6 } \cdot 4 ^ { - 3 } - ( - 2 ) ^ { 3 } - ( \frac { 1 } { 4 } ) ^ { - 1 } ?$$

Evaluate

$$5$$

Solution Steps

Calculate -2 to the power of 6 and get 64.

$$64\times 4^{-3}-\left(-2\right)^{3}-\left(\frac{1}{4}\right)^{-1}$$

Calculate 4 to the power of -3 and get \frac{1}{64}.

$$64\times \frac{1}{64}-\left(-2\right)^{3}-\left(\frac{1}{4}\right)^{-1}$$

Multiply 64 and \frac{1}{64} to get 1.

$$1-\left(-2\right)^{3}-\left(\frac{1}{4}\right)^{-1}$$

Calculate -2 to the power of 3 and get -8.

$$1-\left(-8\right)-\left(\frac{1}{4}\right)^{-1}$$

The opposite of -8 is 8.

$$1+8-\left(\frac{1}{4}\right)^{-1}$$

Add 1 and 8 to get 9.

$$9-\left(\frac{1}{4}\right)^{-1}$$

Calculate \frac{1}{4} to the power of -1 and get 4.

$$9-4$$

Subtract 4 from 9 to get 5.

$$5$$

Factor

$$5$$