Question

If \((9+7\alpha-7\beta)^{20} + (9\alpha+7\beta-7)^{20} + (9\beta+7-7\alpha)^{20} + (14+7\alpha+7\beta)^{20}\) is \(m^{10}\) then the value of m is : (where \(\alpha = \frac{-1+i\sqrt{3}}{2}\) \& \(\beta = \frac{-1-i\sqrt{3}}{2}\))

• A. 50
• B. 49
• C. 46
• D. 48

Hint:

In problems involving sums of powers of expressions with cube roots of unity, look for symmetric relations. Check if the terms are related by multiplication of \(\omega\) or \(\omega^2\). This often leads to a massive simplification using the property \(1+\omega+\omega^2=0\).

Answer 1

The Correct Option is B

Step 1: Understanding the Question: 
The values \(\alpha\) and \(\beta\) are the complex cube roots of unity, usually denoted by \(\omega\) and \(\omega^2\).
We need to simplify a large expression involving these roots. 
Let \(\alpha = \omega\) and \(\beta = \omega^2\). We know the key properties: \(1+\omega+\omega^2=0\) and \(\omega^3=1\). 

Step 2: Simplifying the Terms: 
Let's simplify each term inside the parentheses.
Let \(T_1 = 9+7\alpha-7\beta\), \(T_2 = 9\alpha+7\beta-7\), \(T_3 = 9\beta+7-7\alpha\), and \(T_4 = 14+7\alpha+7\beta\). 

\(T_2 = 9\omega + 7\omega^2 - 7 = 9\omega + 7(-1-\omega) - 7 = 9\omega - 7 - 7\omega - 7 = 2\omega - 14\). 
Let's check the relationship between the terms.
Consider multiplying \(T_2\) by \(\omega\): \(\omega T_2 = \omega(2\omega - 14) = 2\omega^2 - 14\omega\).
Now let's simplify \(T_3\): \(T_3 = 9\beta + 7 - 7\alpha = 9\omega^2 + 7 - 7\omega = 9(-1-\omega) + 7 - 7\omega = -9 - 9\omega + 7 - 7\omega = -2 - 16\omega\). Let's re-evaluate \(\omega T_2 = 2\omega^2 - 14\omega = 2(-1-\omega) - 14\omega = -2 - 2\omega - 14\omega = -2 - 16\omega\).
So, we have found a crucial relation: \(T_3 = \omega T_2\). 
Now let's multiply \(T_2\) by \(\omega^2\): \(\omega^2 T_2 = \omega(\omega T_2) = \omega T_3 = \omega(9\omega^2 + 7 - 7\omega) = 9\omega^3 + 7\omega - 7\omega^2 = 9(1) + 7\omega - 7\omega^2 = 9 + 7\alpha - 7\beta = T_1\). So, another crucial relation is \(T_1 = \omega^2 T_2\). 
Finally, let's simplify \(T_4\): \(T_4 = 14 + 7\alpha + 7\beta = 14 + 7(\alpha+\beta) = 14 + 7(\omega+\omega^2) = 14 + 7(-1) = 7\). 
Step 3: Evaluating the Expression: 
The given expression is \(S = T_1^{20} + T_2^{20} + T_3^{20} + T_4^{20}\). 
Substitute the relations we found: \[ S = (\omega^2 T_2)^{20} + (T_2)^{20} + (\omega T_2)^{20} + T_4^{20} \] \[ S = \omega^{40} T_2^{20} + T_2^{20} + \omega^{20} T_2^{20} + T_4^{20} \] \[ S = T_2^{20} (\omega^{40} + 1 + \omega^{20}) + T_4^{20} \] We simplify the powers of \(\omega\): \(\omega^{40} = (\omega^3)^{13} \cdot \omega = 1^{13} \cdot \omega = \omega\). 
\(\omega^{20} = (\omega^3)^6 \cdot \omega^2 = 1^6 \cdot \omega^2 = \omega^2\). 
So, the term in the parenthesis is \(\omega + 1 + \omega^2\), which is equal to 0. \[ S = T_2^{20} (0) + T_4^{20} = T_4^{20} \] 
Step 4: Final Answer: 
We found that \(T_4 = 7\). So, the expression \(S = 7^{20}\). 
We are given that \(S = m^{10}\). \[ m^{10} = 7^{20} = (7^2)^{10} = 49^{10} \] Therefore, \(m = 49\).