Solve [(9^03+3)*2^3]*3^-2

$$[ ( 9 ^ { 03 } + 3 ) \cdot 2 ^ { 3 } ] \cdot 3 ^ { - 2 }$$

Evaluate

$$\frac{32}{9}\approx 3.555555556$$

Solution Steps

Multiply 0 and 3 to get 0.

$$\left(9^{0}+3\right)\times 2^{3}\times 3^{-2}$$

Calculate 9 to the power of 0 and get 1.

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

Add 1 and 3 to get 4.

$$4\times 2^{3}\times 3^{-2}$$

Calculate 2 to the power of 3 and get 8.

$$4\times 8\times 3^{-2}$$

Multiply 4 and 8 to get 32.

$$32\times 3^{-2}$$

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

$$32\times \frac{1}{9}$$

Multiply 32 and \frac{1}{9} to get \frac{32}{9}.

$$\frac{32}{9}$$

Factor

$$\frac{2 ^ {5}}{3 ^ {2}} = 3\frac{5}{9} = 3.5555555555555554$$