Question

The cube root of a negative integer is
(a) Negative
(b) Complex
(c) Real
(d) Positive

Answer 1

Hint:
Here, we need to find which of the options is correct. Let $$x$$ be a positive integer. Thus, $$-x$$ is a negative integer. First, we will write the cube of the negative integer $$-x$$, and simplify the right hand side such that it is a negative number. Then, we will cube root both sides. Finally, using the equation and the definitions of a negative, complex, real, and positive number, we will find which of the given options is correct.

Complete step by step solution:
We need to find which of the given options is true for the cube root of a negative integer.
An integer is a rational number that is not a fraction.
For example: 1, $$-1$$, 3, $$-7$$, are integers.
Integers can be positive like 1, 3, etc. or negative like $$-1$$.
The cube of a number $$x$$ is given by $$x\times x\times x$$. It can be written using an exponent in the form $$x\times x\times x=x^3$$, where $$x$$ is the base and 3 is the exponent.
Let $$x$$ be a positive integer. Thus, $$-x$$ is a negative integer.
The cube of $$-x$$ can be written as
$$\Rightarrow {\left(-x\right)}^3=\left(-x\right)\times \left(-x\right)\times \left(-x\right)$$
The number $$-x$$ can be written as the product of the negative integer $$-1$$, and the positive integer $$x$$.
Thus, we get
$$\begin{gathered} \Rightarrow {\left(-x\right)}^3=\left(-1\right)\times x\times \left(-1\right)\times x\times \left(-1\right)\times x \\ \Rightarrow {\left(-x\right)}^3={\left(-1\right)}^3\times x\times x\times x \end{gathered}$$
We know that $${\left(-1\right)}^n$$ is equal to 1 if $$n$$ is an even number, and is equal to $$-1$$ if $$n$$ is an odd number.
Therefore, $${\left(-1\right)}^3=-1$$.
The equation becomes
$$\Rightarrow {\left(-x\right)}^3=\left(-1\right)\times x\times x\times x$$
Now, the product of the three positive integers $$x$$ is positive.
The product of the negative integer $$-1$$ and the positive product $$x\times x\times x$$ will be negative.
Therefore, $$\left(-1\right)\times x\times x\times x$$ is a negative integer.
Taking cube root of both sides, we get
$$\Rightarrow -x=\sqrt[3]{\left(-1\right)\times x\times x\times x}$$
Therefore, we can observe that the cube root of the negative integer $$\left(-1\right)\times x\times x\times x$$ is the negative integer $$-x$$.
Thus, option (a) is correct.
We will also check the remaining options because there may be more than one answer.
In a number of the form $$a+bi$$, where $$i=\sqrt{-1}$$, if $$b=0$$, then the number is not complex.
The cube root of the negative integer $$\left(-1\right)\times x\times x\times x$$ is $$-x$$.
The number $$-x$$ can be written as $$-x+0i$$.
Since $$b=0$$, the number $$-x$$ is not complex.
Therefore, the cube root of the negative integer $$\left(-1\right)\times x\times x\times x$$ is not complex.
Thus, option (b) is incorrect.
A real number is any number which is not complex.
We have proved that the cube root of the negative integer $$\left(-1\right)\times x\times x\times x$$ is not complex.
Therefore, the cube root of the negative integer $$\left(-1\right)\times x\times x\times x$$ is a real number.
Thus, option (c) is correct.
We know that a number is either positive or negative, or 0. Any negative number cannot be positive.
We have proved that the cube root of the negative integer $$\left(-1\right)\times x\times x\times x$$ is the negative integer $$-x$$.
Therefore, the cube root of the negative integer $$\left(-1\right)\times x\times x\times x$$ is not a positive number.
Thus, option (d) is incorrect.

We get that options (a) and (c) are correct.

Note:
A complex number is a number which can be written in the form $$a+bi$$, where $$a$$ and $$b$$ are real numbers, and $$i$$ is the imaginary unit. Here, $$i=\sqrt{-1}$$, which is not real.