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------------------------------------------------ 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 monomial of the form
$$a x^b$$
is a perfect cube if both the numerical coefficient $a$ and the variable part $x^b$ represent perfect cubes. In this problem, every monomial is given as $a x^3$, and since the exponent $3$ is a perfect cube (i.e., it is the cube of $1$), we only need to check if the coefficient $a$ is a perfect cube.

A number $n$ is a perfect cube if there exists an integer $k$ such that
$$k^3 = n.$$

Let’s check each option for the coefficient:

1. For the first monomial, the coefficient is $1$. We have
$$1^3 = 1,$$
so $1$ is a perfect cube.

2. For the second monomial, the coefficient is $3$. There is no integer $k$ such that $k^3 = 3$ (since $1^3 = 1$ and $2^3 = 8$).

3. For the third monomial, the coefficient is $6$. Similarly, there is no integer $k$ with $k^3 = 6$.

4. For the fourth monomial, the coefficient is $9$. Again, there is no integer $k$ satisfying $k^3 = 9$.

Since only the coefficient $1$ is a perfect cube, the only monomial that is a perfect cube is
$$1x^3.$$

Thus, the correct answer is:
$$\boxed{1x^3}.$$