Solve [(-3)^{6}div(-3)^3]^2*(-3)^0*(-3)^-4=

$$[ ( - 3 ) ^ { 6 } \div ( - 3 ) ^ { 3 } ] ^ { 2 } \times ( - 3 ) ^ { 0 } \times ( - 3 ) ^ { - 4 } =$$

Evaluate

$$9$$

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.

$$\left(\left(-3\right)^{3}\right)^{2}\left(-3\right)^{0}\left(-3\right)^{-4}$$

To raise a power to another power, multiply the exponents. Multiply 3 and 2 to get 6.

$$\left(-3\right)^{6}\left(-3\right)^{0}\left(-3\right)^{-4}$$

To multiply powers of the same base, add their exponents. Add 6 and 0 to get 6.

$$\left(-3\right)^{6}\left(-3\right)^{-4}$$

To multiply powers of the same base, add their exponents. Add 6 and -4 to get 2.

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

Calculate -3 to the power of 2 and get 9.

$$9$$

Factor

$$3^{2}$$