Solve 2^{0}+(-2)^64^-3-(-2)^3-(1/4)^-1=

$$2 ^ { 0 } + ( - 2 ) ^ { 6 } 4 ^ { - 3 } - ( - 2 ) ^ { 3 } - ( \frac { 1 } { 4 } ) ^ { - 1 } =$$

Evaluate

$$6$$

Solution Steps

Calculate 2 to the power of 0 and get 1.

$$1+\left(-2\right)^{6}\times 4^{-3}-\left(-2\right)^{3}-\left(\frac{1}{4}\right)^{-1}$$

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

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

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

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

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

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

Add 1 and 1 to get 2.

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

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

$$2-\left(-8\right)-\left(\frac{1}{4}\right)^{-1}$$

The opposite of -8 is 8.

$$2+8-\left(\frac{1}{4}\right)^{-1}$$

Add 2 and 8 to get 10.

$$10-\left(\frac{1}{4}\right)^{-1}$$

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

$$10-4$$

Subtract 4 from 10 to get 6.

$$6$$

Factor

$$2\times 3$$