Solve 20*(-2)^3*(1/5)^3-32*(-2)^2*(1/5)^2

$$20 \times ( - 2 ) ^ { 3 } \times ( \frac { 1 } { 5 } ) ^ { 3 } - 32 \times ( - 2 ) ^ { 2 } \times ( \frac { 1 } { 5 } ) ^ { 2 }$$

Evaluate

$$-\frac{32}{5}=-6.4$$

Solution Steps

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

$$20\left(-8\right)\times \left(\frac{1}{5}\right)^{3}-32\left(-2\right)^{2}\times \left(\frac{1}{5}\right)^{2}$$

Multiply 20 and -8 to get -160.

$$-160\times \left(\frac{1}{5}\right)^{3}-32\left(-2\right)^{2}\times \left(\frac{1}{5}\right)^{2}$$

Calculate \frac{1}{5} to the power of 3 and get \frac{1}{125}.

$$-160\times \frac{1}{125}-32\left(-2\right)^{2}\times \left(\frac{1}{5}\right)^{2}$$

Multiply -160 and \frac{1}{125} to get \frac{-160}{125}.

$$\frac{-160}{125}-32\left(-2\right)^{2}\times \left(\frac{1}{5}\right)^{2}$$

Reduce the fraction \frac{-160}{125} to lowest terms by extracting and canceling out 5.

$$-\frac{32}{25}-32\left(-2\right)^{2}\times \left(\frac{1}{5}\right)^{2}$$

Calculate -2 to the power of 2 and get 4.

$$-\frac{32}{25}-32\times 4\times \left(\frac{1}{5}\right)^{2}$$

Multiply 32 and 4 to get 128.

$$-\frac{32}{25}-128\times \left(\frac{1}{5}\right)^{2}$$

Calculate \frac{1}{5} to the power of 2 and get \frac{1}{25}.

$$-\frac{32}{25}-128\times \frac{1}{25}$$

Multiply 128 and \frac{1}{25} to get \frac{128}{25}.

$$-\frac{32}{25}-\frac{128}{25}$$

Since -\frac{32}{25} and \frac{128}{25} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-32-128}{25}$$

Subtract 128 from -32 to get -160.

$$\frac{-160}{25}$$

Reduce the fraction \frac{-160}{25} to lowest terms by extracting and canceling out 5.

$$-\frac{32}{5}$$

Factor

$$-\frac{32}{5} = -6\frac{2}{5} = -6.4$$