Verified Answer :
- Express the decimal \(0.2\) as the fraction \(\frac{1}{5}\).
- Calculate the cube of the fraction: \(\left(\frac{1}{5}\right)^3 = \frac{1^3}{5^3} = \frac{1}{125}\).
- Convert the fraction back to decimal form: \(\frac{1}{125} = 0.008\).
- The final result is \(\boxed{0.008}\).
Explanation
- Analysis of the Expression
The expression we need to evaluate is \((0.2)^3\). This involves raising the decimal \(0.2\) to the power of \(3\), which means multiplying the number by itself three times. We can solve this either by working directly with decimals or by converting the decimal to a fraction to simplify the process.
- Fractional Calculation
A helpful approach is to convert the decimal \(0.2\) into its fractional form. We know that \(0.2 = \frac{2}{10}\). Simplifying this fraction by dividing both the numerator and the denominator by \(2\), we get:
\[0.2 = \frac{1}{5}\]
Now, we can rewrite the original expression as:
\[(0.2)^3 = \left(\frac{1}{5}\right)^3\]
Using the exponent rule for fractions, \(\left(\frac{a}{b}\right)^n = \frac{a^n}{b^n}\), we have:
\[\left(\frac{1}{5}\right)^3 = \frac{1^3}{5^3} = \frac{1 \times 1 \times 1}{5 \times 5 \times 5}\]
Calculating the denominator, \(5 \times 5 \times 5 = 25 \times 5 = 125\). Thus:
\[\left(\frac{1}{5}\right)^3 = \frac{1}{125}\]
- Final Evaluation
To express the final result as a decimal, we can divide \(1\) by \(125\). Alternatively, we can multiply the numerator and denominator by \(8\) to get a power of \(10\) in the denominator:
\[\frac{1}{125} = \frac{1 \times 8}{125 \times 8} = \frac{8}{1000} = 0.008\]
Alternatively, multiplying directly in decimal form: \(0.2 \times 0.2 = 0.04\), and \(0.04 \times 0.2 = 0.008\). Both methods yield the same result. The value of the expression is \(\boxed{0.008}\).
Examples
This type of calculation is very useful in everyday life, for instance, when calculating volumes of small cubes or scaling down items. If you are designing a miniature model and need to scale down the dimensions by a factor of 0.2 (i.e., 1/5th), the volume of the original object will decrease by a factor of \((0.2)^3 = 0.008\). This shows how even small changes in linear dimensions lead to significant changes in three-dimensional volume!