Answer :

To determine which monomial is a perfect cube, let's first understand what a perfect cube is. A number (or monomial) is considered a perfect cube if it can be expressed as the cube of another number.

For the given monomials [tex]\( 1x^3 \)[/tex], [tex]\( 3x^3 \)[/tex], [tex]\( 6x^3 \)[/tex], and [tex]\( 9x^3 \)[/tex], we need to check the coefficients (1, 3, 6, and 9) to see if any of them are perfect cubes.

Step-by-step Solution:

1. Identify the Coefficients:
- The given monomials have coefficients 1, 3, 6, and 9.

2. Check Each Coefficient:
- To determine if a coefficient is a perfect cube, check if there exists an integer whose cube equals the coefficient:
- For 1: [tex]\( 1^3 = 1 \)[/tex] (This is a perfect cube.)
- For 3: [tex]\( \sqrt[3]{3} \)[/tex] is not an integer.
- For 6: [tex]\( \sqrt[3]{6} \)[/tex] is not an integer.
- For 9: [tex]\( \sqrt[3]{9} \)[/tex] is not an integer.

3. Conclusion:
- Among the coefficients 1, 3, 6, and 9, only the number 1 is a perfect cube because [tex]\( 1^3 = 1 \)[/tex].

Therefore, the monomial [tex]\( 1x^3 \)[/tex] is a perfect cube because its coefficient, 1, is a perfect cube.

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