Answer :
To determine which monomial is a perfect cube, we'll focus on the coefficients of the given monomials. A monomial is a perfect cube if both its coefficient and its variable part are perfect cubes.
Let's examine the coefficients of the monomials presented in the options:
1. [tex]\(1 \cdot x^3\)[/tex]
2. [tex]\(3 \cdot x^3\)[/tex]
3. [tex]\(6 \cdot x^3\)[/tex]
4. [tex]\(9 \cdot x^3\)[/tex]
For these monomials, note that the variable part [tex]\(x^3\)[/tex] is already a perfect cube because [tex]\(x^3 = (x)^3\)[/tex].
Now, let's determine which coefficients are perfect cubes:
- 1: Since [tex]\(1 = 1^3\)[/tex], 1 is a perfect cube.
- 3: There is no whole number [tex]\(a\)[/tex] such that [tex]\(a^3 = 3\)[/tex].
- 6: There is no whole number [tex]\(a\)[/tex] such that [tex]\(a^3 = 6\)[/tex].
- 9: There is no whole number [tex]\(a\)[/tex] such that [tex]\(a^3 = 9\)[/tex].
From this analysis, only the coefficient of [tex]\(1\)[/tex] is a perfect cube. Therefore, the monomial [tex]\(1 \cdot x^3\)[/tex] is a perfect cube.
The monomial that is a perfect cube is [tex]\(1 \cdot x^3\)[/tex].
Let's examine the coefficients of the monomials presented in the options:
1. [tex]\(1 \cdot x^3\)[/tex]
2. [tex]\(3 \cdot x^3\)[/tex]
3. [tex]\(6 \cdot x^3\)[/tex]
4. [tex]\(9 \cdot x^3\)[/tex]
For these monomials, note that the variable part [tex]\(x^3\)[/tex] is already a perfect cube because [tex]\(x^3 = (x)^3\)[/tex].
Now, let's determine which coefficients are perfect cubes:
- 1: Since [tex]\(1 = 1^3\)[/tex], 1 is a perfect cube.
- 3: There is no whole number [tex]\(a\)[/tex] such that [tex]\(a^3 = 3\)[/tex].
- 6: There is no whole number [tex]\(a\)[/tex] such that [tex]\(a^3 = 6\)[/tex].
- 9: There is no whole number [tex]\(a\)[/tex] such that [tex]\(a^3 = 9\)[/tex].
From this analysis, only the coefficient of [tex]\(1\)[/tex] is a perfect cube. Therefore, the monomial [tex]\(1 \cdot x^3\)[/tex] is a perfect cube.
The monomial that is a perfect cube is [tex]\(1 \cdot x^3\)[/tex].
