Solve [(-2)^{6}:(-2)^3]^3*(-2)*(-2)^-4=^*

$$[ ( - 2 ) ^ { 6 } : ( - 2 ) ^ { 3 } ] ^ { 3 } \cdot ( - 2 ) \cdot ( - 2 ) ^ { - 4 } = ^ { * }$$

Evaluate

$$64$$

Solution Steps

To divide powers of the same base, subtract the denominator's exponent from the numerator's exponent. Subtract 3 from 6 to get 3.

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

To raise a power to another power, multiply the exponents. Multiply 3 and 3 to get 9.

$$\left(-2\right)^{9}\left(-2\right)\left(-2\right)^{-4}$$

To multiply powers of the same base, add their exponents. Add 9 and 1 to get 10.

$$\left(-2\right)^{10}\left(-2\right)^{-4}$$

To multiply powers of the same base, add their exponents. Add 10 and -4 to get 6.

$$\left(-2\right)^{6}$$

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

$$64$$

Factor

$$2^{6}$$