Factorizing \(x^3 - 1\) Using Cube Roots of Unity

The question asks us to factorize the expression \(x^3 - 1\), given that \(\omega\) is one of the cube roots of unity.

Understanding Cube Roots of Unity

The cube roots of unity are the solutions to the equation \(x^3 = 1\). These are complex numbers that, when cubed, equal 1. There are exactly three cube roots of unity in the complex number system. They are:

• 1 (the real root)
• \(\omega\) (a complex root)
• \(\omega^2\) (the other complex root, which is the square of \(\omega\))

These roots have important properties:

• The sum of the cube roots of unity is zero: \(1 + \omega + \omega^2 = 0\).
• The product of the cube roots of unity is one: \(1 \cdot \omega \cdot \omega^2 = \omega^3 = 1\).

Connecting Roots to Factorization

For any polynomial \(P(x)\), if \(r\) is a root of the polynomial (meaning \(P(r) = 0\)), then \((x - r)\) is a factor of the polynomial. A polynomial of degree \(n\) can be factored into \(n\) linear factors of the form \((x - r_i)\), where \(r_i\) are its roots.

In our case, the expression is \(x^3 - 1\). We want to find the roots of the equation \(x^3 - 1 = 0\), which is the same as \(x^3 = 1\).

As we discussed, the roots of \(x^3 = 1\) are the cube roots of unity: \(1\), \(\omega\), and \(\omega^2\).

So, the roots of \(x^3 - 1\) are \(r_1 = 1\), \(r_2 = \omega\), and \(r_3 = \omega^2\).

Factoring \(x^3 - 1\)

Since the roots of \(x^3 - 1\) are \(1\), \(\omega\), and \(\omega^2\), the factors are \((x - 1)\), \((x - \omega)\), and \((x - \omega^2)\).

Therefore, the expression \(x^3 - 1\) can be factored as the product of these factors:

\(x^3 - 1 = (x - 1)(x - \omega)(x - \omega^2)\)

Verifying the Factorization (Optional)

We can expand the factored form to verify:

\((x - \omega)(x - \omega^2) = x^2 - x\omega^2 - x\omega + \omega \cdot \omega^2 = x^2 - x(\omega + \omega^2) + \omega^3\)

Using the properties of cube roots of unity, we know that \(\omega + \omega^2 = -1\) (from \(1 + \omega + \omega^2 = 0\)) and \(\omega^3 = 1\).

Substituting these values:

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

Now, multiply by the remaining factor \((x - 1)\):

\((x - 1)(x^2 + x + 1)\)

This is a standard algebraic identity for the difference of cubes:

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

Setting \(a = x\) and \(b = 1\), we get:

\(x^3 - 1^3 = (x - 1)(x^2 + x + 1)\)

So, the factorization \((x - 1)(x - \omega)(x - \omega^2)\) is correct because \((x - \omega)(x - \omega^2) = x^2 + x + 1\).

Comparing with Options

Let's compare our result with the given options:

The factorization that matches our derivation is Option 2.

Revision Table: Key Concepts

Additional Information on Cube Roots of Unity

The complex cube roots of unity, \(\omega\) and \(\omega^2\), can be represented in polar form:

• \(1 = e^{i 0} = \cos(0) + i \sin(0)\)
• \(\omega = e^{i 2\pi/3} = \cos(2\pi/3) + i \sin(2\pi/3) = -\frac{1}{2} + i\frac{\sqrt{3}}{2}\)
• \(\omega^2 = e^{i 4\pi/3} = \cos(4\pi/3) + i \sin(4\pi/3) = -\frac{1}{2} - i\frac{\sqrt{3}}{2}\)

Geometrically, the cube roots of unity are points on the unit circle in the complex plane, located at angles \(0\), \(2\pi/3\), and \(4\pi/3\) radians from the positive real axis. They form the vertices of an equilateral triangle inscribed in the unit circle.

The property \(1 + \omega + \omega^2 = 0\) is very useful in simplifying expressions involving powers of \(\omega\).