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Answer :
A perfect cube is an expression that can be written in the form
[tex]$$
n^3,
$$[/tex]
where [tex]$n$[/tex] is an integer.
The given monomials have the form
[tex]$$
(\text{coefficient})x^3.
$$[/tex]
Since the variable part [tex]$x^3$[/tex] is already a cube (because [tex]$x^3=(x)^3$[/tex]), we only need to check which coefficient is a perfect cube.
A coefficient [tex]$c$[/tex] is a perfect cube if there exists an integer [tex]$n$[/tex] such that
[tex]$$
n^3 = c.
$$[/tex]
Let’s examine each coefficient:
1. For the monomial [tex]$1x^3$[/tex]:
The coefficient is [tex]$1$[/tex]. We know that
[tex]$$
1^3 = 1.
$$[/tex]
Hence, [tex]$1$[/tex] is a perfect cube.
2. For the monomial [tex]$3x^3$[/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]$6x^3$[/tex]:
The coefficient is [tex]$6$[/tex], and there is no integer [tex]$n$[/tex] satisfying [tex]$n^3 = 6$[/tex], so [tex]$6$[/tex] is not a perfect cube.
4. For the monomial [tex]$9x^3$[/tex]:
The coefficient is [tex]$9$[/tex], and no integer [tex]$n$[/tex] satisfies [tex]$n^3 = 9$[/tex], so [tex]$9$[/tex] is not a perfect cube.
Since only the coefficient [tex]$1$[/tex] meets the condition for a perfect cube, the monomial
[tex]$$
1x^3
$$[/tex]
is the perfect cube.
Thus, the answer is:
[tex]$$
\boxed{1x^3}.
$$[/tex]
[tex]$$
n^3,
$$[/tex]
where [tex]$n$[/tex] is an integer.
The given monomials have the form
[tex]$$
(\text{coefficient})x^3.
$$[/tex]
Since the variable part [tex]$x^3$[/tex] is already a cube (because [tex]$x^3=(x)^3$[/tex]), we only need to check which coefficient is a perfect cube.
A coefficient [tex]$c$[/tex] is a perfect cube if there exists an integer [tex]$n$[/tex] such that
[tex]$$
n^3 = c.
$$[/tex]
Let’s examine each coefficient:
1. For the monomial [tex]$1x^3$[/tex]:
The coefficient is [tex]$1$[/tex]. We know that
[tex]$$
1^3 = 1.
$$[/tex]
Hence, [tex]$1$[/tex] is a perfect cube.
2. For the monomial [tex]$3x^3$[/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]$6x^3$[/tex]:
The coefficient is [tex]$6$[/tex], and there is no integer [tex]$n$[/tex] satisfying [tex]$n^3 = 6$[/tex], so [tex]$6$[/tex] is not a perfect cube.
4. For the monomial [tex]$9x^3$[/tex]:
The coefficient is [tex]$9$[/tex], and no integer [tex]$n$[/tex] satisfies [tex]$n^3 = 9$[/tex], so [tex]$9$[/tex] is not a perfect cube.
Since only the coefficient [tex]$1$[/tex] meets the condition for 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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