If `alphaandbeta` are the roots of the equation `x^(2)-6x+12=0` and the value of `(alpha-2)^(24)-((beta-6)^(8))/(alpha^(8))+1` is `4^(a)`, then the value of a is ______.

Explanation

To solve the problem step by step, we will follow the instructions provided in the video transcript.

### Step 1: Identify the roots of the quadratic equation
The given equation is:
\[ x^2 - 6x + 12 = 0 \]

We can find the roots using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1 \), \( b = -6 \), and \( c = 12 \).

Calculating the discriminant:
\[ b^2 - 4ac = (-6)^2 - 4 \cdot 1 \cdot 12 = 36 - 48 = -12 \]
Since the discriminant is negative, the roots are complex.

Now, substituting into the quadratic formula:
\[ x = \frac{6 \pm \sqrt{-12}}{2} = \frac{6 \pm 2i\sqrt{3}}{2} = 3 \pm i\sqrt{3} \]
Let:
\[ \alpha = 3 + i\sqrt{3}, \quad \beta = 3 - i\sqrt{3} \]

### Step 2: Substitute \(\beta\) into the expression
We need to evaluate:
\[ \frac{(\alpha - 2)^{24} - (\beta - 6)^{8}}{\alpha^{8}} + 1 \]

First, calculate \(\alpha - 2\) and \(\beta - 6\):
\[ \alpha - 2 = (3 + i\sqrt{3}) - 2 = 1 + i\sqrt{3} \]
\[ \beta - 6 = (3 - i\sqrt{3}) - 6 = -3 - i\sqrt{3} \]

### Step 3: Calculate the powers
Now we calculate:
\[ (\alpha - 2)^{24} = (1 + i\sqrt{3})^{24} \]
Using polar form:
\[ 1 + i\sqrt{3} = 2 \left( \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) \right) \]
Thus,
\[ (1 + i\sqrt{3})^{24} = 2^{24} \left( \cos\left(8\pi\right) + i\sin\left(8\pi\right) \right) = 2^{24} \cdot 1 = 2^{24} \]

Next, calculate \((\beta - 6)^{8}\):
\[ \beta - 6 = -3 - i\sqrt{3} = 2 \left( \cos\left(\frac{4\pi}{3}\right) + i\sin\left(\frac{4\pi}{3}\right) \right) \]
Thus,
\[ (-3 - i\sqrt{3})^{8} = 2^{8} \left( \cos\left(\frac{32\pi}{3}\right) + i\sin\left(\frac{32\pi}{3}\right) \right) = 2^{8} \cdot 1 = 2^{8} \]

### Step 4: Calculate \(\alpha^{8}\)
Now, calculate \(\alpha^{8}\):
\[ \alpha = 3 + i\sqrt{3} = 2 \left( \cos\left(\frac{\pi}{3}\right) + i\sin\left(\frac{\pi}{3}\right) \right) \]
Thus,
\[ \alpha^{8} = 2^{8} \left( \cos\left(\frac{8\pi}{3}\right) + i\sin\left(\frac{8\pi}{3}\right) \right) = 2^{8} \cdot 1 = 2^{8} \]

### Step 5: Substitute into the main expression
Now substituting back into the expression:
\[ \frac{2^{24} - 2^{8}}{2^{8}} + 1 = \frac{2^{24} - 2^{8}}{2^{8}} + 1 = 2^{16} - 1 + 1 = 2^{16} \]

### Step 6: Set equal to \(4^a\)
We know that:
\[ 2^{16} = 4^8 \]
Thus, we have:
\[ 4^a = 4^8 \implies a = 8 \]

### Final Answer
The value of \( a \) is:
\[ \boxed{8} \]