Solve 3^{0}*(-3)^3*3^-2+(-2)^3*(-1)^2=

$$3 ^ { 0 } \cdot ( - 3 ) ^ { 3 } \cdot 3 ^ { - 2 } + ( - 2 ) ^ { 3 } \cdot ( - 1 ) ^ { 2 } =$$

Evaluate

$$-11$$

Solution Steps

To multiply powers of the same base, add their exponents. Add 0 and -2 to get -2.

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

Calculate 3 to the power of -2 and get \frac{1}{9}.

$$\frac{1}{9}\left(-3\right)^{3}+\left(-2\right)^{3}\left(-1\right)^{2}$$

Calculate -3 to the power of 3 and get -27.

$$\frac{1}{9}\left(-27\right)+\left(-2\right)^{3}\left(-1\right)^{2}$$

Multiply \frac{1}{9} and -27 to get -3.

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

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

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

Calculate -1 to the power of 2 and get 1.

$$-3-8$$

Subtract 8 from -3 to get -11.

$$-11$$

Factor

$$-11$$