Answer :
To determine which monomial is a perfect cube, we need to consider both the coefficient and the variable part of each option. A monomial is considered a perfect cube when both its coefficient is a perfect cube and the exponent of its variable is a multiple of 3.
Let's examine each option:
1. [tex]$1 x^3$[/tex]:
- The coefficient is 1. Since [tex]\(1^3 = 1\)[/tex], the number 1 is a perfect cube.
- The exponent of [tex]\(x\)[/tex] is 3, which is a multiple of 3.
- Therefore, [tex]\(1 x^3\)[/tex] is a perfect cube.
2. [tex]$3 x^3$[/tex]:
- The coefficient is 3. There is no integer [tex]\( n \)[/tex] such that [tex]\(n^3 = 3\)[/tex], so 3 is not a perfect cube.
- Even though the exponent of [tex]\(x\)[/tex] is 3, the coefficient must also be a perfect cube, which it isn't.
- Therefore, [tex]\(3 x^3\)[/tex] is not a perfect cube.
3. [tex]$6 x^3$[/tex]:
- The coefficient is 6. There is no integer [tex]\( n \)[/tex] such that [tex]\(n^3 = 6\)[/tex], so 6 is not a perfect cube.
- The exponent of [tex]\(x\)[/tex] is 3, but since the coefficient is not a perfect cube, [tex]\(6 x^3\)[/tex] is not a perfect cube.
4. [tex]$9 x^3$[/tex]:
- The coefficient is 9. There is no integer [tex]\( n \)[/tex] such that [tex]\(n^3 = 9\)[/tex], so 9 is not a perfect cube.
- Although the exponent of [tex]\(x\)[/tex] is 3, the coefficient needs to be a perfect cube, which it isn't, so [tex]\(9 x^3\)[/tex] is not a perfect cube.
Based on this analysis, the only monomial that is indeed a perfect cube is [tex]\(1 x^3\)[/tex].
Let's examine each option:
1. [tex]$1 x^3$[/tex]:
- The coefficient is 1. Since [tex]\(1^3 = 1\)[/tex], the number 1 is a perfect cube.
- The exponent of [tex]\(x\)[/tex] is 3, which is a multiple of 3.
- Therefore, [tex]\(1 x^3\)[/tex] is a perfect cube.
2. [tex]$3 x^3$[/tex]:
- The coefficient is 3. There is no integer [tex]\( n \)[/tex] such that [tex]\(n^3 = 3\)[/tex], so 3 is not a perfect cube.
- Even though the exponent of [tex]\(x\)[/tex] is 3, the coefficient must also be a perfect cube, which it isn't.
- Therefore, [tex]\(3 x^3\)[/tex] is not a perfect cube.
3. [tex]$6 x^3$[/tex]:
- The coefficient is 6. There is no integer [tex]\( n \)[/tex] such that [tex]\(n^3 = 6\)[/tex], so 6 is not a perfect cube.
- The exponent of [tex]\(x\)[/tex] is 3, but since the coefficient is not a perfect cube, [tex]\(6 x^3\)[/tex] is not a perfect cube.
4. [tex]$9 x^3$[/tex]:
- The coefficient is 9. There is no integer [tex]\( n \)[/tex] such that [tex]\(n^3 = 9\)[/tex], so 9 is not a perfect cube.
- Although the exponent of [tex]\(x\)[/tex] is 3, the coefficient needs to be a perfect cube, which it isn't, so [tex]\(9 x^3\)[/tex] is not a perfect cube.
Based on this analysis, the only monomial that is indeed a perfect cube is [tex]\(1 x^3\)[/tex].
