Answer :

Sure! To determine which monomial is a perfect cube, let's consider all the given options and understand what makes a monomial a perfect cube.

A monomial is a perfect cube if both the coefficient (the numerical part) and the variable part (like [tex]\(x^3\)[/tex]) are perfect cubes. In all given options, the variable part is [tex]\(x^3\)[/tex], which is already a cube. So, we only need to focus on the numerical coefficients.

Let's analyze each option:

1. Option: [tex]\(1x^3\)[/tex]
- The numerical coefficient is 1.
- 1 is a perfect cube because [tex]\(1 = 1^3\)[/tex].

2. Option: [tex]\(3x^3\)[/tex]
- The numerical coefficient is 3.
- 3 is not a perfect cube because there is no integer [tex]\(n\)[/tex] such that [tex]\(n^3 = 3\)[/tex].

3. Option: [tex]\(6x^3\)[/tex]
- The numerical coefficient is 6.
- 6 is not a perfect cube because there is no integer [tex]\(n\)[/tex] such that [tex]\(n^3 = 6\)[/tex].

4. Option: [tex]\(9x^3\)[/tex]
- The numerical coefficient is 9.
- 9 is not a perfect cube because no integer [tex]\(n\)[/tex] satisfies [tex]\(n^3 = 9\)[/tex]. In fact, 9 is [tex]\(3^2\)[/tex], a perfect square but not a cube.

So, among the given options, only [tex]\(1x^3\)[/tex] has a coefficient that is a perfect cube. Therefore, the monomial [tex]\(1x^3\)[/tex] is a perfect cube.

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Rewritten by : Barada