Which of the following is a perfect cube?
216
Identifying Perfect Cubes
A perfect cube is an integer that is the cube of another integer. In other words, it is a number that can be obtained by multiplying an integer by itself three times.
To determine if a number is a perfect cube, we can use the method of prime factorization. A number is a perfect cube if, in its prime factorization, the exponent of each prime factor is a multiple of 3.
Checking the Given Options
Option 1: 270
Let's find the prime factorization of 270:
\(270 = 2 \times 135\)
\(135 = 3 \times 45\)
\(45 = 3 \times 15\)
\(15 = 3 \times 5\)
So, \(270 = 2 \times 3 \times 3 \times 3 \times 5 = 2^1 \times 3^3 \times 5^1\).
The exponents in the prime factorization are 1, 3, and 1. Since not all exponents are multiples of 3 (1 and 1 are not multiples of 3), 270 is not a perfect cube.
Option 2: 2700
Let's find the prime factorization of 2700:
\(2700 = 27 \times 100\)
\(27 = 3^3\)
\(100 = 10^2 = (2 \times 5)^2 = 2^2 \times 5^2\)
So, \(2700 = 3^3 \times 2^2 \times 5^2 = 2^2 \times 3^3 \times 5^2\).
The exponents in the prime factorization are 2, 3, and 2. Since not all exponents are multiples of 3 (2 and 2 are not multiples of 3), 2700 is not a perfect cube.
Option 3: 256
Let's find the prime factorization of 256:
\(256 = 2 \times 128\)
\(128 = 2 \times 64\)
\(64 = 2 \times 32\)
\(32 = 2 \times 16\)
\(16 = 2 \times 8\)
\(8 = 2 \times 4\)
\(4 = 2 \times 2\)
So, \(256 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 = 2^8\).
The exponent in the prime factorization is 8. Since 8 is not a multiple of 3, 256 is not a perfect cube.
Option 4: 216
Let's find the prime factorization of 216:
\(216 = 2 \times 108\)
\(108 = 2 \times 54\)
\(54 = 2 \times 27\)
\(27 = 3 \times 9\)
\(9 = 3 \times 3\)
So, \(216 = 2 \times 2 \times 2 \times 3 \times 3 \times 3 = 2^3 \times 3^3\).
The exponents in the prime factorization are 3 and 3. Since both exponents are multiples of 3, 216 is a perfect cube. We can also see that \(216 = 6^3\).
Summary of Prime Factorizations
| Number | Prime Factorization | Exponents | Is a Perfect Cube? |
|---|---|---|---|
| 270 | \(2^1 \times 3^3 \times 5^1\) | 1, 3, 1 | No |
| 2700 | \(2^2 \times 3^3 \times 5^2\) | 2, 3, 2 | No |
| 256 | \(2^8\) | 8 | No |
| 216 | \(2^3 \times 3^3\) | 3, 3 | Yes |
Based on the analysis, only 216 has exponents that are all multiples of 3 in its prime factorization.
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