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 used as a cube root, because cubing it multiplies each variable’s exponent by 3. For example, cubing x² gives x⁶, whose exponent is divisible by 3. The cube root’s exponents do not themselves have to be divisible by 3; it is the exponents in the resulting perfect cube that are divisible by 3.

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