Find the value of w⁴ + w⁶ + w⁸, if w is a complex cube root of unity.

To find the value of \( w^4 + w^6 + w^8 \) where \( w \) is a complex cube root of unity, we first need to understand the properties of \( w \).

Properties of Cube Roots of Unity

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

• \( w = e^{2\pi i / 3} \)
• \( w^2 = e^{4\pi i / 3} \)
• \( w^3 = 1 \)

From this, we know that \( w^3 = 1 \) and thus \( w^n \) can be simplified based on its exponent modulo 3.

Simplifying the Exponents

Now, let's simplify each term:

• \( w^4 = w^{3+1} = w^1 = w \)
• \( w^6 = w^{3 \cdot 2} = (w^3)^2 = 1^2 = 1 \)
• \( w^8 = w^{3 \cdot 2 + 2} = w^2 \)

Calculating the Sum

Now we can substitute these values back into the expression:

\( w^4 + w^6 + w^8 = w + 1 + w^2 \)

Using the Sum of Roots

Since \( 1 + w + w^2 = 0 \) (the sum of the cube roots of unity), we can rearrange this to find:

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

Final Result

Thus, we have:

\( w^4 + w^6 + w^8 = -1 + 1 = 0 \)

The final value is 0.