Is 108 a perfect cube? This question often arises in mathematics, especially when discussing the properties of numbers and their cubes. In this article, we will explore the concept of perfect cubes, examine the properties of the number 108, and determine whether it is indeed a perfect cube.
A perfect cube is a number that can be expressed as the cube of an integer. In other words, if a number \( n \) is a perfect cube, then there exists an integer \( m \) such that \( n = m^3 \). To determine if 108 is a perfect cube, we need to find an integer \( m \) that, when cubed, equals 108.
Let’s start by analyzing the prime factorization of 108. By breaking down 108 into its prime factors, we can gain insight into its properties. The prime factorization of 108 is \( 2^2 \times 3^3 \). This means that 108 can be expressed as the product of 2 raised to the power of 2 and 3 raised to the power of 3.
Now, let’s consider the cube of an integer. If \( m \) is an integer, then \( m^3 \) will have prime factors raised to the power of 3. In our case, we are looking for an integer \( m \) such that \( m^3 = 2^2 \times 3^3 \).
To find such an integer, we can look at the exponents of the prime factors. Since the exponent of 2 in the prime factorization of 108 is 2, and the exponent of 3 is 3, we need to find an integer \( m \) such that the exponents of its prime factors are also 2 and 3, respectively.
Let’s consider the prime factorization of \( m^3 \). Since \( m^3 \) is a perfect cube, its prime factors will be raised to the power of 3. Therefore, the prime factorization of \( m^3 \) will be of the form \( p^3 \times q^3 \), where \( p \) and \( q \) are prime numbers.
Comparing the prime factorization of 108 with the prime factorization of \( m^3 \), we can see that the prime factorization of 108 cannot be expressed as \( p^3 \times q^3 \). The prime factorization of 108 has a 2 raised to the power of 2 and a 3 raised to the power of 3, whereas the prime factorization of \( m^3 \) would require a 2 raised to the power of 3 and a 3 raised to the power of 3.
Therefore, we can conclude that 108 is not a perfect cube. This is because there is no integer \( m \) such that \( m^3 = 108 \). The number 108 has prime factors raised to the power of 2 and 3, respectively, which means it cannot be expressed as the cube of an integer.
