Evaluate: (1-omega+omega2)5 + (1+omega-omega2)5, where omega is a cube root of unity.

Question

ScanSolve AI Tutor · Accepted Answer

To evaluate the expression, we use the properties of the cube roots of unity, where $1 + \omega + \omega^2 = 0$ .

Simplify the first term $(1-\omega+\omega^2)^5$ .
From the property $1 + \omega + \omega^2 = 0$ , we know that $1 + \omega^2 = -\omega$ .
Substitute this into the first term:
$(1-\omega+\omega^2)^5 = ((1+\omega^2) - \omega)^5 = (-\omega - \omega)^5 = (-2\omega)^5$
Simplify the second term $(1+\omega-\omega^2)^5$ .
From the property $1 + \omega + \omega^2 = 0$ , we know that $1 + \omega = -\omega^2$ .
Substitute this into the second term:
$(1+\omega-\omega^2)^5 = ((1+\omega) - \omega^2)^5 = (-\omega^2 - \omega^2)^5 = (-2\omega^2)^5$
Add the simplified terms.
$(-2\omega)^5 + (-2\omega^2)^5 = (-2)^5 \omega^5 + (-2)^5 (\omega^2)^5$ $= -32 \omega^5 - 32 \omega^{10}$
Factor out $-32$ :
$= -32 (\omega^5 + \omega^{10})$
Simplify the powers of $\omega$ .
We know that $\omega^3 = 1$ .
So, $\omega^5 = \omega^3 \cdot \omega^2 = 1 \cdot \omega^2 = \omega^2$ .
And $\omega^{10} = \omega^9 \cdot \omega = (\omega^3)^3 \cdot \omega = 1^3 \cdot \omega = \omega$ .
Substitute the simplified powers back into the expression.
$-32 (\omega^2 + \omega)$
From the property $1 + \omega + \omega^2 = 0$ , we know that $\omega + \omega^2 = -1$ .
$-32 (-1) = 32$
The final answer is $\text{(d) } 32$ .

Answer