Question

miguel stated that any monomial can be a cube root. sylvia disagreed and said that a monomial cube root must have exponents divisible by 3. who is correct, and why?

• miguel is correct. any monomial can be a perfect cube root because, when it is cubed, the variables will have exponents divisible by 3.
• sylvia is correct. in order for the cube to have exponents that are divisible by 3, the cube root has to be divisible by 3.
• sylvia is correct. any variable term, when it is cubed, always has an exponent divisible by 3.
• miguel is correct. sylvia confused the perfect square root with the perfect cube root.

Explanation:

Define the concept of a cube root

A monomial \(B\) is a cube root of another monomial \(A\) if cubing \(B\) results in \(A\):

$$ B^3 = A $$

Thus, any monomial \(B\) can serve as a cube root.

Apply exponent rules to cubing a monomial

Using the Exponent Rules knowledge point

$$ (c \cdot x^n)^3 = c^3 \cdot x^{3n} $$

Since \(n\) is an integer, the exponent of the resulting perfect cube \(3n\) is always divisible by 3. However, the starting monomial \(B = c \cdot x^n\) can have any integer exponent \(n\).

Evaluate Miguel and Sylvia's claims

Miguel states any monomial can be a cube root. This is correct because we can cube any monomial \(x^n\) to get a perfect cube \(x^{3n}\).
Sylvia claims a cube root must have exponents divisible by 3. This is incorrect; it is the perfect cube itself that must have exponents divisible by 3, not its cube root.

Match with the correct option

Miguel is correct. Any monomial can be a perfect cube root because, when it is cubed, the variables will have exponents divisible by 3.
This matches the first option.

Answer:

• Miguel is correct. Any monomial can be a perfect cube root because, when it is cubed, the variables will have exponents divisible by 3. (Correct answer)
• Sylvia is correct. In order for the cube to have exponents that are divisible by 3, the cube root has to be divisible by 3.
• Sylvia is correct. Any variable term, when it is cubed, always has an exponent divisible by 3.
• Miguel is correct. Sylvia confused the perfect square root with the perfect cube root.