Answer :
To determine which monomial is a perfect cube, we need to examine the coefficients of the given monomials. A monomial is considered a perfect cube if its coefficient is a perfect cube number, and the exponent of the variable is a multiple of 3.
Here are the monomials we are evaluating:
1. [tex]\( 1x^3 \)[/tex]
2. [tex]\( 3x^3 \)[/tex]
3. [tex]\( 6x^3 \)[/tex]
4. [tex]\( 9x^3 \)[/tex]
We will check each coefficient to see if it is a perfect cube:
1. [tex]\( 1 \)[/tex]:
- The cube root of 1 is [tex]\( 1 \)[/tex], since [tex]\( 1^3 = 1 \)[/tex].
- Therefore, 1 is a perfect cube.
2. [tex]\( 3 \)[/tex]:
- The cube root of 3 is not an integer, since there is no integer [tex]\( n \)[/tex] such that [tex]\( n^3 = 3 \)[/tex].
- Thus, 3 is not a perfect cube.
3. [tex]\( 6 \)[/tex]:
- Similarly, the cube root of 6 is not an integer, so there is no integer [tex]\( n \)[/tex] such that [tex]\( n^3 = 6 \)[/tex].
- Therefore, 6 is not a perfect cube.
4. [tex]\( 9 \)[/tex]:
- The cube root of 9 is not an integer, since there is no integer [tex]\( n \)[/tex] such that [tex]\( n^3 = 9 \)[/tex].
- Thus, 9 is not a perfect cube.
After evaluating each coefficient, we find that the only coefficient that is a perfect cube is 1. Additionally, since the exponent of the variable [tex]\( x \)[/tex] in all monomials is 3, they all fit the criterion for being a perfect cube in terms of the variable part.
Therefore, the monomial [tex]\( 1x^3 \)[/tex] is the only one that is a perfect cube.
Here are the monomials we are evaluating:
1. [tex]\( 1x^3 \)[/tex]
2. [tex]\( 3x^3 \)[/tex]
3. [tex]\( 6x^3 \)[/tex]
4. [tex]\( 9x^3 \)[/tex]
We will check each coefficient to see if it is a perfect cube:
1. [tex]\( 1 \)[/tex]:
- The cube root of 1 is [tex]\( 1 \)[/tex], since [tex]\( 1^3 = 1 \)[/tex].
- Therefore, 1 is a perfect cube.
2. [tex]\( 3 \)[/tex]:
- The cube root of 3 is not an integer, since there is no integer [tex]\( n \)[/tex] such that [tex]\( n^3 = 3 \)[/tex].
- Thus, 3 is not a perfect cube.
3. [tex]\( 6 \)[/tex]:
- Similarly, the cube root of 6 is not an integer, so there is no integer [tex]\( n \)[/tex] such that [tex]\( n^3 = 6 \)[/tex].
- Therefore, 6 is not a perfect cube.
4. [tex]\( 9 \)[/tex]:
- The cube root of 9 is not an integer, since there is no integer [tex]\( n \)[/tex] such that [tex]\( n^3 = 9 \)[/tex].
- Thus, 9 is not a perfect cube.
After evaluating each coefficient, we find that the only coefficient that is a perfect cube is 1. Additionally, since the exponent of the variable [tex]\( x \)[/tex] in all monomials is 3, they all fit the criterion for being a perfect cube in terms of the variable part.
Therefore, the monomial [tex]\( 1x^3 \)[/tex] is the only one that is a perfect cube.
