Answer :

To determine which monomial is a perfect cube, let's analyze each option. A monomial is a perfect cube if it can be written as [tex]\(a^3 \cdot x^3\)[/tex] where [tex]\(a\)[/tex] is a whole number.

Let's check:

1. [tex]\(1x^3\)[/tex]
- The coefficient is 1.
- [tex]\(1 = 1^3\)[/tex], so 1 is a perfect cube.
- Therefore, [tex]\(1x^3 = (1x)^3\)[/tex].
- [tex]\(1x^3\)[/tex] is a perfect cube.

2. [tex]\(3x^3\)[/tex]
- The coefficient is 3.
- 3 is not a perfect cube (since [tex]\(a^3 = 3\)[/tex] doesn't hold for any integer [tex]\(a\)[/tex]).
- Therefore, [tex]\(3x^3\)[/tex] is not a perfect cube.

3. [tex]\(6x^3\)[/tex]
- The coefficient is 6.
- 6 is not a perfect cube (since [tex]\(a^3 = 6\)[/tex] doesn't hold for any integer [tex]\(a\)[/tex]).
- Therefore, [tex]\(6x^3\)[/tex] is not a perfect cube.

4. [tex]\(9x^3\)[/tex]
- The coefficient is 9.
- 9 is not a perfect cube (since [tex]\(a^3 = 9\)[/tex] doesn't hold for any integer [tex]\(a\)[/tex]).
- Therefore, [tex]\(9x^3\)[/tex] is not a perfect cube.

From our analysis, we can see that the monomial [tex]\(1x^3\)[/tex] is indeed a perfect cube. Thus, the answer is [tex]\(1x^3\)[/tex].

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