Answer :
To determine which monomial is a perfect cube, we need to consider both the coefficient and the variable part of the monomial.
A monomial is a perfect cube if both the coefficient and the powers of the variables are perfect cubes. Let's look at the given options:
1. [tex]\(1 x^3\)[/tex]
2. [tex]\(3 x^3\)[/tex]
3. [tex]\(6 x^3\)[/tex]
4. [tex]\(9 x^3\)[/tex]
Step 1: Check if the coefficient is a perfect cube.
- A number is a perfect cube if it's the result of an integer multiplied by itself twice (e.g., [tex]\(a^3\)[/tex]).
- Here are the coefficients: 1, 3, 6, and 9.
Let's check each:
- 1: The cube root of 1 is 1, which is an integer. So, 1 is a perfect cube.
- 3: The cube root of 3 is not an integer, so 3 is not a perfect cube.
- 6: The cube root of 6 is not an integer, so 6 is not a perfect cube.
- 9: The cube root of 9 is not an integer, so 9 is not a perfect cube.
Step 2: Check the variable part.
Since all terms have [tex]\(x^3\)[/tex], and 3 is a multiple of 3, the power of [tex]\(x\)[/tex] is a perfect cube.
Conclusion:
The monomial [tex]\(1 x^3\)[/tex] is a perfect cube because both the coefficient 1 and the exponent of [tex]\(x^3\)[/tex] meet the perfect cube condition. Therefore, [tex]\(1 x^3\)[/tex] is the monomial that is a perfect cube.
A monomial is a perfect cube if both the coefficient and the powers of the variables are perfect cubes. Let's look at the given options:
1. [tex]\(1 x^3\)[/tex]
2. [tex]\(3 x^3\)[/tex]
3. [tex]\(6 x^3\)[/tex]
4. [tex]\(9 x^3\)[/tex]
Step 1: Check if the coefficient is a perfect cube.
- A number is a perfect cube if it's the result of an integer multiplied by itself twice (e.g., [tex]\(a^3\)[/tex]).
- Here are the coefficients: 1, 3, 6, and 9.
Let's check each:
- 1: The cube root of 1 is 1, which is an integer. So, 1 is a perfect cube.
- 3: The cube root of 3 is not an integer, so 3 is not a perfect cube.
- 6: The cube root of 6 is not an integer, so 6 is not a perfect cube.
- 9: The cube root of 9 is not an integer, so 9 is not a perfect cube.
Step 2: Check the variable part.
Since all terms have [tex]\(x^3\)[/tex], and 3 is a multiple of 3, the power of [tex]\(x\)[/tex] is a perfect cube.
Conclusion:
The monomial [tex]\(1 x^3\)[/tex] is a perfect cube because both the coefficient 1 and the exponent of [tex]\(x^3\)[/tex] meet the perfect cube condition. Therefore, [tex]\(1 x^3\)[/tex] is the monomial that is a perfect cube.
