Solve (10^3-10^-2)*(3^3-3^1)*1/3=

$$( { 10 }^{ 3 } - { 10 }^{ -2 } ) \times ( { 3 }^{ 3 } - { 3 }^{ 1 } ) \times \frac{ 1 }{ 3 } =$$

Evaluate

$$\frac{199998}{25}=7999.92$$

Solution Steps

Calculate 10 to the power of 3 and get 1000.

$$\left(1000-10^{-2}\right)\left(3^{3}-3^{1}\right)\times \frac{1}{3}$$

Calculate 10 to the power of -2 and get \frac{1}{100}.

$$\left(1000-\frac{1}{100}\right)\left(3^{3}-3^{1}\right)\times \frac{1}{3}$$

Subtract \frac{1}{100} from 1000 to get \frac{99999}{100}.

$$\frac{99999}{100}\left(3^{3}-3^{1}\right)\times \frac{1}{3}$$

Calculate 3 to the power of 3 and get 27.

$$\frac{99999}{100}\left(27-3^{1}\right)\times \frac{1}{3}$$

Calculate 3 to the power of 1 and get 3.

$$\frac{99999}{100}\left(27-3\right)\times \frac{1}{3}$$

Subtract 3 from 27 to get 24.

$$\frac{99999}{100}\times 24\times \frac{1}{3}$$

Multiply \frac{99999}{100} and 24 to get \frac{599994}{25}.

$$\frac{599994}{25}\times \frac{1}{3}$$

Multiply \frac{599994}{25} and \frac{1}{3} to get \frac{199998}{25}.

$$\frac{199998}{25}$$

Factor

$$\frac{2 \cdot 3 ^ {2} \cdot 41 \cdot 271}{5 ^ {2}} = 7999\frac{23}{25} = 7999.92$$