Answer :

To find which monomial is a perfect cube, we need to understand what makes a monomial a perfect cube. A monomial is a perfect cube when both the numerical coefficient and the powers of the variables are perfect cubes.

Let's examine each option:

1. [tex]\(1x^3\)[/tex]:
- The coefficient is 1. Since [tex]\(1 = 1^3\)[/tex], the coefficient is a perfect cube.
- The variable [tex]\(x^3\)[/tex] already has an exponent that is a multiple of 3.
- Thus, [tex]\(1x^3\)[/tex] is a perfect cube.

2. [tex]\(3x^3\)[/tex]:
- The coefficient is 3, which is not a perfect cube because there is no integer [tex]\(a\)[/tex] such that [tex]\(a^3 = 3\)[/tex].
- Even though the variable [tex]\(x^3\)[/tex] has an exponent that is a multiple of 3, the coefficient makes the whole term not a perfect cube.

3. [tex]\(6x^3\)[/tex]:
- The coefficient is 6, which is not a perfect cube because there is no integer [tex]\(a\)[/tex] such that [tex]\(a^3 = 6\)[/tex].
- Hence, [tex]\(6x^3\)[/tex] is not a perfect cube.

4. [tex]\(9x^3\)[/tex]:
- The coefficient is 9, which is not a perfect cube because there is no integer [tex]\(a\)[/tex] such that [tex]\(a^3 = 9\)[/tex].
- Therefore, [tex]\(9x^3\)[/tex] is also not a perfect cube.

To conclude, the monomial [tex]\(1x^3\)[/tex] is the only perfect cube among the given options because both its coefficient and the exponent on the variable [tex]\(x\)[/tex] are perfect cubes.

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