Which choices are real numbers? Check all that apply.

A. [tex]$(-32)^{1/6}$[/tex]

B. [tex]$(-10)^{1/4}$[/tex]

C. [tex]$(-4)^{1/3}$[/tex]

D. [tex]$(-16)^{1/5}$[/tex]



Answer :

To determine which of the given choices result in real numbers, we need to analyze each one individually, focusing on the properties of exponents and roots.

### Choice A: [tex]\((-32)^{1/6}\)[/tex]
- Here, we are taking the sixth root of [tex]\(-32\)[/tex].
- When raising a negative number to a fractional power with an even denominator, the result is an imaginary number because even roots of negative numbers are not defined in the real number system.
- Therefore, [tex]\((-32)^{1/6}\)[/tex] is not a real number.

### Choice B: [tex]\((-10)^{1/4}\)[/tex]
- Here, we are taking the fourth root of [tex]\(-10\)[/tex].
- Similar to Choice A, we are dealing with an even root.
- A negative number raised to a power with an even denominator results in an imaginary number.
- Therefore, [tex]\((-10)^{1/4}\)[/tex] is not a real number.

### Choice C: [tex]\((-4)^{1/3}\)[/tex]
- Here, we are taking the cube root of [tex]\(-4\)[/tex].
- When raising a negative number to a fractional power with an odd denominator, the result is a real number because odd roots of negative numbers are defined in the real number system.
- Therefore, [tex]\((-4)^{1/3}\)[/tex] is a real number.

### Choice D: [tex]\((-16)^{1/5}\)[/tex]
- Here, we are taking the fifth root of [tex]\(-16\)[/tex].
- Similar to Choice C, we are dealing with an odd root.
- An odd root of a negative number is a real number.
- Therefore, [tex]\((-16)^{1/5}\)[/tex] is a real number.

### Summary
Based on the analysis, the choices that result in real numbers are:

- [tex]\( \boxed{C} \)[/tex] [tex]\((-4)^{1/3}\)[/tex]
- [tex]\( \boxed{D} \)[/tex] [tex]\((-16)^{1/5}\)[/tex]

So the correct answers are:
[tex]\[ \boxed{\text{C and D}} \][/tex]

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