State the total number of complex roots, the possible number of real and imaginary roots and the possible rational roots for each equation. Then find all roots. $$ x^{3}-1=0 $$ a \# of complex roots: 3 Possible \# of real roots: 3 or 1 Possible \# of imaginary roots: 2 or 0 Possible rational roots: $$ \pm 1 $$ $$ \text { Roots: } \pm 1, \frac{-1+i \sqrt{3}}{2} $$ b $$ 1, \frac{-1+i \sqrt{3}}{2}, \frac{-1-i \sqrt{3}}{2} \text { \# of complex roots: } 3 $$ Possible \# of real roots: 3 or 1 Possible \# of imaginary roots: 2 or 0 Possible rational roots: $$ \pm 1 $$ Roots: c \# of complex roots: 3 Possible \# of real roots: 3 or 1 Possible \# of imaginary roots: 2 or 0 Possible rational roots: $$ \pm 1 $$ $$ \text { Roots: }-1, \frac{-1+i \sqrt{2}}{5}, \frac{-1-i \sqrt{2}}{5} $$ d \# of complex roots: 3 Possible \# of real roots: 3 or 1 Possible \# of imaginary roots: 2 or 0 Possible rational roots: $$ \pm 1 $$ Roots: $$ -1, \frac{i \sqrt{3}}{2}, \frac{-i \sqrt{3}}{2} $$

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Let's analyze the equation $$x^3 - 1 = 0$$ step by step to determine the total number of complex roots, possible real and imaginary roots, possible rational roots, and then find all the roots.

### Step 1: Total Number of Complex Roots
The degree of the polynomial $$x^3 - 1$$ is 3. By the Fundamental Theorem of Algebra, a polynomial of degree $$n$$ has exactly $$n$$ roots in the complex number system (counting multiplicities). Therefore, the total number of complex roots is:
- **Total number of complex roots: 3**

### Step 2: Possible Number of Real and Imaginary Roots
For a cubic polynomial, the possible distributions of real and imaginary roots can be:
- 3 real roots (all roots are real)
- 1 real root and 2 complex conjugate roots (which are imaginary)

Thus, the possible number of real and imaginary roots is:
- **Possible number of real roots: 3 or 1**
- **Possible number of imaginary roots: 2 or 0**

### Step 3: Possible Rational Roots
Using the Rational Root Theorem, the possible rational roots of the polynomial $$x^3 - 1$$ can be determined from the factors of the constant term (-1) and the leading coefficient (1). The possible rational roots are:
- **Possible rational roots: $$ \pm 1 $$**

### Step 4: Finding All Roots
To find the roots of the equation $$x^3 - 1 = 0$$, we can factor it as follows:
$$
x^3 - 1 = (x - 1)(x^2 + x + 1) = 0
$$
This gives us one real root:
- $$x = 1$$

Next, we need to solve the quadratic equation $$x^2 + x + 1 = 0$$ using the quadratic formula:
$$
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
$$
where $$a = 1$$, $$b = 1$$, and $$c = 1$$:
$$
x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2}
$$
Thus, the roots of the equation $$x^3 - 1 = 0$$ are:
- **Roots: $$1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}$$**

### Summary
- **Total number of complex roots: 3**
- **Possible number of real roots: 3 or 1**
- **Possible number of imaginary roots: 2 or 0**
- **Possible rational roots: $$ \pm 1 $$**
- **Roots: $$1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}$$**

This completes the analysis and solution for the equation $$x^3 - 1 = 0$$.
The equation $$x^3 - 1 = 0$$ has: - **3 complex roots** - **Possible real roots: 3 or 1** - **Possible imaginary roots: 2 or 0** - **Possible rational roots: $$ \pm 1 $$** - **Roots: $$1, \frac{-1 + i\sqrt{3}}{2}, \frac{-1 - i\sqrt{3}}{2}$$**

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