Answer :

To determine which monomial is a perfect cube, we note that a monomial of the form

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

is a perfect cube if both the coefficient [tex]$c$[/tex] and the variable part [tex]$x^3$[/tex] are perfect cubes. Since

[tex]$$
x^3 = (x)^3,
$$[/tex]

the variable part is already a perfect cube. Therefore, it remains to check whether the numerical coefficient [tex]$c$[/tex] is a perfect cube.

A number is a perfect cube if it can be written as some integer [tex]$k$[/tex] cubed, i.e., [tex]$c = k^3$[/tex]. Let’s analyze the coefficient for each option:

1. For [tex]$1x^3$[/tex], the coefficient is [tex]$1$[/tex]. We have

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

so [tex]$1$[/tex] is a perfect cube.

2. For [tex]$3x^3$[/tex], the coefficient is [tex]$3$[/tex]. There is no integer [tex]$k$[/tex] such that [tex]$k^3 = 3$[/tex], so [tex]$3$[/tex] is not a perfect cube.

3. For [tex]$6x^3$[/tex], the coefficient is [tex]$6$[/tex]. There is no integer [tex]$k$[/tex] such that [tex]$k^3 = 6$[/tex], so [tex]$6$[/tex] is not a perfect cube.

4. For [tex]$9x^3$[/tex], the coefficient is [tex]$9$[/tex]. There is no integer [tex]$k$[/tex] such that [tex]$k^3 = 9$[/tex], so [tex]$9$[/tex] is not a perfect cube.

Since the only coefficient that is a perfect cube is [tex]$1$[/tex], it follows that the monomial

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

is a perfect cube.

Thus, the answer is: [tex]$$1x^3.$$[/tex]

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