Answer :
- Check if the variable part is a perfect cube.
- Check if the coefficient of each monomial is a perfect cube.
- Identify the monomial with both the variable part and the coefficient as perfect cubes.
- The monomial $1x^3$ is a perfect cube because $1 = 1^3$ and $x^3 = (x)^3$. The final answer is $\boxed{1 x^3}$.
### Explanation
1. Understanding the Problem
We are given four monomials: $1x^3$, $3x^3$, $6x^3$, and $9x^3$. We need to determine which one is a perfect cube. A perfect cube is a number or expression that can be written as the cube of another number or expression. For a monomial to be a perfect cube, both the coefficient and the variable part must be perfect cubes.
2. Checking the Variable Part
The variable part $x^3$ is already a perfect cube since it is $(x)^3$. So, we only need to check if the coefficient of each monomial is a perfect cube.
3. Analyzing the Coefficient of $1x^3$
For $1x^3$, the coefficient is 1, and $1 = 1^3$, so 1 is a perfect cube. Therefore, $1x^3$ is a perfect cube.
4. Analyzing the Coefficient of $3x^3$
For $3x^3$, the coefficient is 3. Since 3 is not a perfect cube (i.e., there is no integer $n$ such that $n^3 = 3$), $3x^3$ is not a perfect cube.
5. Analyzing the Coefficient of $6x^3$
For $6x^3$, the coefficient is 6. Since 6 is not a perfect cube, $6x^3$ is not a perfect cube.
6. Analyzing the Coefficient of $9x^3$
For $9x^3$, the coefficient is 9. Since 9 is not a perfect cube, $9x^3$ is not a perfect cube.
7. Conclusion
Therefore, the only monomial with a perfect cube coefficient is $1x^3$.
### Examples
Perfect cubes are useful in various applications, such as calculating the volume of a cube. For example, if you have a cube-shaped box with sides of length $x$, its volume is $x^3$. Understanding perfect cubes helps in determining the side length if you know the volume is a perfect cube. This concept extends to other areas like engineering and architecture where cubic relationships are common.
- Check if the coefficient of each monomial is a perfect cube.
- Identify the monomial with both the variable part and the coefficient as perfect cubes.
- The monomial $1x^3$ is a perfect cube because $1 = 1^3$ and $x^3 = (x)^3$. The final answer is $\boxed{1 x^3}$.
### Explanation
1. Understanding the Problem
We are given four monomials: $1x^3$, $3x^3$, $6x^3$, and $9x^3$. We need to determine which one is a perfect cube. A perfect cube is a number or expression that can be written as the cube of another number or expression. For a monomial to be a perfect cube, both the coefficient and the variable part must be perfect cubes.
2. Checking the Variable Part
The variable part $x^3$ is already a perfect cube since it is $(x)^3$. So, we only need to check if the coefficient of each monomial is a perfect cube.
3. Analyzing the Coefficient of $1x^3$
For $1x^3$, the coefficient is 1, and $1 = 1^3$, so 1 is a perfect cube. Therefore, $1x^3$ is a perfect cube.
4. Analyzing the Coefficient of $3x^3$
For $3x^3$, the coefficient is 3. Since 3 is not a perfect cube (i.e., there is no integer $n$ such that $n^3 = 3$), $3x^3$ is not a perfect cube.
5. Analyzing the Coefficient of $6x^3$
For $6x^3$, the coefficient is 6. Since 6 is not a perfect cube, $6x^3$ is not a perfect cube.
6. Analyzing the Coefficient of $9x^3$
For $9x^3$, the coefficient is 9. Since 9 is not a perfect cube, $9x^3$ is not a perfect cube.
7. Conclusion
Therefore, the only monomial with a perfect cube coefficient is $1x^3$.
### Examples
Perfect cubes are useful in various applications, such as calculating the volume of a cube. For example, if you have a cube-shaped box with sides of length $x$, its volume is $x^3$. Understanding perfect cubes helps in determining the side length if you know the volume is a perfect cube. This concept extends to other areas like engineering and architecture where cubic relationships are common.
