To solve the equation \(x^3 - 1 = 0\), we first rewrite it in a more convenient form:

\[x^3 - 1 = 0\]

We can factor this equation using the difference of cubes formula:

\[a^3 - b^3 = (a - b)(a^2 + ab + b^2)\]

Here, \(a = x\) and \(b = 1\). So, we have:

\[x^3 - 1 = (x - 1)(x^2 + x \cdot 1 + 1^2) = (x - 1)(x^2 + x + 1)\]

Now, we set each factor equal to zero to find the roots:

1. \(x - 1 = 0\)
2. \(x^2 + x + 1 = 0\)

Solving the first equation:

\[x - 1 = 0 \Rightarrow x = 1\]

Now, solving the quadratic equation \(x^2 + x + 1 = 0\):

We use 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 two complex roots are:

\[x = \frac{-1 + i\sqrt{3}}{2}\]
\[x = \frac{-1 - i\sqrt{3}}{2}\]

So, the three roots of the equation \(x^3 - 1 = 0\) are:

1. \(x = 1\)
2. \(x = \frac{-1 + i\sqrt{3}}{2}\)
3. \(x = \frac{-1 - i\sqrt{3}}{2}\)

Would you like further details on this solution or have any other questions?

Here are 5 questions you might want to ask next:
1. How do you solve quadratic equations?
2. What is the geometric interpretation of complex roots?
3. Can you explain the difference of cubes formula?
4. How do you factor polynomials?
5. What are some real-world applications of finding roots of equations?

**Tip:** When solving polynomial equations, factoring is a powerful tool that simplifies the problem, making it easier to find all roots.

Explore how to solve the equation x^3 - 1 = 0 step-by-step, including factoring using the difference of cubes formula and solving the resulting quadratic equation. Learn about polynomial equations and complex roots. Ideal for advanced high school students and anyone interested in polynomial equations.

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