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]
[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]
