Understanding Cube Roots of Unity

The question asks for the value of the expression \(xy + yz + zx\), where \(x, y,\) and \(z\) represent the cube roots of unity.

The cube roots of unity are the solutions to the equation \(t^3 = 1\). These roots are:

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

Key Properties of Cube Roots of Unity

These roots have important properties that are often used in algebra:

• Sum of Roots: The sum of the cube roots of unity is always zero. \(1 + \omega + \omega^2 = 0\)
• Product of Roots: The product of the cube roots is one. \(1 \times \omega \times \omega^2 = \omega^3 = 1\)
• Powers of \(\omega\): Any power of \(\omega\) can be simplified using \(\omega^3 = 1\). For example, \(\omega^4 = \omega^3 \cdot \omega = 1 \cdot \omega = \omega\), and \(\omega^5 = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2\).

Calculating \(xy + yz + zx\)

Let's assign the cube roots of unity to \(x, y,\) and \(z\). A common assignment is:

• \(x = 1\)
• \(y = \omega\)
• \(z = \omega^2\)

Now, we can calculate each term in the expression \(xy + yz + zx\):

• \(xy = 1 \times \omega = \omega\)
• \(yz = \omega \times \omega^2 = \omega^3\)
• \(zx = \omega^2 \times 1 = \omega^2\)

Using the property \(\omega^3 = 1\), the term \(yz\) becomes \(1\). So, the expression is:

From the property of the sum of cube roots of unity (\(1 + \omega + \omega^2 = 0\)), we get:

Alternative Method Using Vieta's Formulas

The equation \(t^3 = 1\) can be written as \(t^3 - 1 = 0\). This is a cubic equation of the form \(at^3 + bt^2 + ct + d = 0\), where \(a=1\), \(b=0\), \(c=0\), and \(d=-1\).

If \(x, y,\) and \(z\) are the roots of this equation, Vieta's formulas relate the coefficients to the sums and products of the roots:

• Sum of roots: \(x + y + z = -b/a\)
• Sum of the product of roots taken two at a time: \(xy + yz + zx = c/a\)
• Product of roots: \(xyz = -d/a\)

Using the coefficients from \(t^3 - 1 = 0\) (\(a=1, b=0, c=0\)):

Both methods confirm that the value of the expression is \(0\). The assignment of \(x, y, z\) to \(1, \omega, \omega^2\) does not change the result due to the symmetric nature of the expression and the properties of the roots.