Answer :

A monomial is a perfect cube if its numerical coefficient is the cube of an integer. In other words, a number [tex]$a$[/tex] is a perfect cube if there exists an integer [tex]$n$[/tex] such that

[tex]$$
n^3 = a.
$$[/tex]

We will check each coefficient:

1. For the monomial [tex]$1x^3$[/tex], the coefficient is [tex]$1$[/tex]. Since

[tex]$$
1^3 = 1,
$$[/tex]

the number [tex]$1$[/tex] is a perfect cube.

2. For the monomial [tex]$3x^3$[/tex], the coefficient is [tex]$3$[/tex]. The cube root of [tex]$3$[/tex] is not an integer; indeed, there is no integer [tex]$n$[/tex] satisfying [tex]$n^3 = 3$[/tex].

3. For the monomial [tex]$6x^3$[/tex], the coefficient is [tex]$6$[/tex]. Similarly, there is no integer [tex]$n$[/tex] such that [tex]$n^3 = 6$[/tex].

4. For the monomial [tex]$9x^3$[/tex], the coefficient is [tex]$9$[/tex]. Again, there is no integer [tex]$n$[/tex] with [tex]$n^3 = 9$[/tex].

Since only [tex]$1$[/tex] is a perfect cube, the monomial

[tex]$$
1x^3
$$[/tex]

is the perfect cube.

Thus, the answer is:

[tex]$$
\boxed{1x^3}.
$$[/tex]

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