Answer :
To solve the equation [tex]\(\frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} = \omega\)[/tex], let's proceed step by step:
1. Multiply Both Sides by the Denominator:
We start by eliminating the fraction by multiplying both sides of the equation by the denominator [tex]\(b + c\omega + a\omega^2\)[/tex]:
[tex]\[ a + b\omega + c\omega^2 = \omega \cdot (b + c\omega + a\omega^2) \][/tex]
2. Distribute [tex]\(\omega\)[/tex] on the Right Side:
Distribute [tex]\(\omega\)[/tex] through the terms inside the parentheses:
[tex]\[ a + b\omega + c\omega^2 = \omega b + \omega c\omega + \omega a\omega^2 \][/tex]
3. Simplify the Terms:
Since [tex]\(\omega \cdot \omega = \omega^2\)[/tex] and [tex]\(\omega \cdot \omega^2 = \omega^3\)[/tex], we can rewrite the equation as:
[tex]\[ a + b\omega + c\omega^2 = \omega b + c\omega^2 \omega + a\omega^3 \][/tex]
Given that [tex]\(\omega^3 = 1\)[/tex] (a fundamental property of cubic roots of unity), simplify further:
[tex]\[ \omega \cdot \omega = \omega^2 \quad \text{and} \quad \omega \cdot \omega^2 = \omega^3 = 1 \][/tex]
Therefore, the equation becomes:
[tex]\[ a + b\omega + c\omega^2 = b\omega + c\omega^2 + a \][/tex]
4. Observe the Simplified Equation:
We see that the left and right sides of the equation are exactly equal:
[tex]\[ a + b\omega + c\omega^2 = a + b\omega + c\omega^2 \][/tex]
This equality holds true for any values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] because both sides of the equation are identical.
### Conclusion
The original equation [tex]\(\frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} = \omega\)[/tex] holds true for any values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]. Therefore, the solution is that:
The equation holds for any [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex].
1. Multiply Both Sides by the Denominator:
We start by eliminating the fraction by multiplying both sides of the equation by the denominator [tex]\(b + c\omega + a\omega^2\)[/tex]:
[tex]\[ a + b\omega + c\omega^2 = \omega \cdot (b + c\omega + a\omega^2) \][/tex]
2. Distribute [tex]\(\omega\)[/tex] on the Right Side:
Distribute [tex]\(\omega\)[/tex] through the terms inside the parentheses:
[tex]\[ a + b\omega + c\omega^2 = \omega b + \omega c\omega + \omega a\omega^2 \][/tex]
3. Simplify the Terms:
Since [tex]\(\omega \cdot \omega = \omega^2\)[/tex] and [tex]\(\omega \cdot \omega^2 = \omega^3\)[/tex], we can rewrite the equation as:
[tex]\[ a + b\omega + c\omega^2 = \omega b + c\omega^2 \omega + a\omega^3 \][/tex]
Given that [tex]\(\omega^3 = 1\)[/tex] (a fundamental property of cubic roots of unity), simplify further:
[tex]\[ \omega \cdot \omega = \omega^2 \quad \text{and} \quad \omega \cdot \omega^2 = \omega^3 = 1 \][/tex]
Therefore, the equation becomes:
[tex]\[ a + b\omega + c\omega^2 = b\omega + c\omega^2 + a \][/tex]
4. Observe the Simplified Equation:
We see that the left and right sides of the equation are exactly equal:
[tex]\[ a + b\omega + c\omega^2 = a + b\omega + c\omega^2 \][/tex]
This equality holds true for any values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex] because both sides of the equation are identical.
### Conclusion
The original equation [tex]\(\frac{a + b\omega + c\omega^2}{b + c\omega + a\omega^2} = \omega\)[/tex] holds true for any values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]. Therefore, the solution is that:
The equation holds for any [tex]\(a\)[/tex], [tex]\(b\)[/tex], [tex]\(c\)[/tex].