Which expression is equivalent to $21 \sqrt[3]{15}-9 \sqrt[3]{15}$ ?


A12
B$30 \sqrt[3]{15}$
C$12 \sqrt[3]{5}$
D$12 \sqrt[3]{15}$

Verified Answer :

  • Identify the common radical part, \(\sqrt[3]{15}\), in both terms.
  • Use the distributive property to group the coefficients: \((21 - 9) \sqrt[3]{15}\).
  • Perform the subtraction: \(21 - 9 = 12\).
  • State the final simplified form as \(\boxed{12 \sqrt[3]{15}}\).

Explanation

  1. Problem Analysis

The objective is to simplify the algebraic expression \(21 \sqrt[3]{15} - 9 \sqrt[3]{15}\). This expression involves two terms, each containing the same radical component, \(\sqrt[3]{15}\), multiplied by different numerical coefficients: 21 and 9. Recognizing that both terms share a common factor allows us to use the distributive property to simplify the expression efficiently.

  1. Applying the Distributive Property

The distributive property for real numbers states that for any real numbers \(a, b,\) and \(c,\) the expression \(ac - bc\) can be written as \((a - b)c\). In our case, we let \(a = 21\), \(b = 9\), and \(c = \sqrt[3]{15}\). Therefore, the expression \(21 \sqrt[3]{15} - 9 \sqrt[3]{15}\) can be rewritten as:
\[(21 - 9) \sqrt[3]{15}\]

  1. Calculating the Difference and Conclusion

Now, we calculate the difference between the two coefficients inside the parentheses:
\[21 - 9 = 12\]
By substituting this result back into our expression, we obtain:
\[12 \sqrt[3]{15}\]
This simplified form represents the original expression. Comparing this to the given choices, we see that it matches the final option.

Examples

In daily life, this algebraic skill is similar to grouping like items when counting inventory. For example, if you have 21 bundles of 15 items each and you remove 9 of those same bundles, you are left with \(21 - 9 = 12\) bundles, each still containing 15 items. This principle is fundamental in fields ranging from physics to finance, where managing identical units across complex terms is a daily necessity for accurate accounting and calculation.