Question

3. Why do cube roots of negative numbers exist in the set of real numbers,

Answer 1

I'd go with

C. The square root of a number, positive or negative, is always positive.

However, the cube of a negative number is negative.

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That's not totally true and it's a bit off the point.  The square root of a negative number isn't a positive number, it's a positive number times i.

Even that's not totally true. We need to distinguish between a square root and the principal square root function.  What's true is every non zero complex number has two square roots, negations of each other. $\sqrt{x}$, the radical sign applied to a real number, is a function, because we define it to be by choosing one of the square roots as the principal square root.  

The right answer would be more like

E. The square of a non-zero real number is always a positive real, so considering the inverse, negative reals don't have real square roots.  The cube of a  real number has the same sign as the original number, so again considering the inverse, negative reals do have real cube roots.

Answer 2

Answer:

  C. The square root of a number, positive or negative, is always positive. However, the cube of a negative number is negative.

Step-by-step explanation:

Of the choices offered, the one shown above is the only one that remotely makes any sense. Its description of square roots is erroneous, however.

The square root of a number n is the solution to the equation ...

  x² = n

The function f(x) = x² has a range that only includes non-negative numbers, so the square root of a negative number does not exist. (Your answer choice says it is positive. It is actually non-existent among real numbers.)

On the other hand, the cube root of a number n is the solution to the equation ...

  x³ = n

Since f(x) = x³ has a range that includes all real numbers, the cube root of n exists for all real numbers. (It is negative for negative values of n.)

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Comment on answer wording

The answer wording would be correct if the word "root" were left out.