College

Which monomial is a perfect cube?

A. [tex]1x^3[/tex]
B. [tex]3x^3[/tex]
C. [tex]6x^3[/tex]
D. [tex]9x^3[/tex]

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]