Answer :

To determine which monomial is a perfect cube, we analyze the monomials of the form

[tex]$$
a x^3,
$$[/tex]

where [tex]$x^3$[/tex] is already a perfect cube. Therefore, the monomial will be a perfect cube if and only if the numerical coefficient [tex]$a$[/tex] is also a perfect cube.

Let's check the given options:

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

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

it is a perfect cube.

2. For the monomial [tex]$$3 x^3,$$[/tex] the coefficient is [tex]$$3.$$[/tex] There is no integer [tex]$n$[/tex] such that

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

so [tex]$3$[/tex] is not a perfect cube.

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

[tex]$$
n^3 = 6,
$$[/tex]

so [tex]$6$[/tex] is not a perfect cube.

4. For the monomial [tex]$$9 x^3,$$[/tex] the coefficient is [tex]$$9.$$[/tex] There is no integer [tex]$n$[/tex] such that

[tex]$$
n^3 = 9,
$$[/tex]

so [tex]$9$[/tex] is not a perfect cube.

Since the only coefficient that is a perfect cube is [tex]$$1,$$[/tex] the monomial [tex]$$\mathbf{1 x^3}$$[/tex] is the perfect cube.

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