The sum of the coefficients of all even degree terms is x in the expansion of : `(x+sqrt(x^(3)-1))^(6)+(x-sqrt(x^(3)-1))^(6), (xgt1)` is equal to

Options

A. 29
B. 32
C. 26
D. 24

Answer

To find the sum of the coefficients of all even degree terms in the expression \((x + \sqrt{x^3 - 1})^6 + (x - \sqrt{x^3 - 1})^6\), we can follow these steps:

### Step 1: Expand the Expression
We start with the expression:
\[ (x + \sqrt{x^3 - 1})^6 + (x - \sqrt{x^3 - 1})^6 \]
Using the Binomial Theorem, we can expand both terms separately.

### Step 2: Apply the Binomial Theorem
The Binomial Theorem states:
\[ (a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k \]
For \( (x + \sqrt{x^3 - 1})^6 \):
\[ = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} (\sqrt{x^3 - 1})^k \]
For \( (x - \sqrt{x^3 - 1})^6 \):
\[ = \sum_{k=0}^{6} \binom{6}{k} x^{6-k} (-\sqrt{x^3 - 1})^k \]

### Step 3: Combine the Expansions
When we add these two expansions, the odd powers of \(\sqrt{x^3 - 1}\) will cancel out:
\[ (x + \sqrt{x^3 - 1})^6 + (x - \sqrt{x^3 - 1})^6 = 2 \sum_{k \text{ even}} \binom{6}{k} x^{6-k} (x^3 - 1)^{k/2} \]
This leaves us with only the even terms.

### Step 4: Identify Even Terms
The even terms occur when \(k\) is even, specifically \(k = 0, 2, 4, 6\):
- For \(k = 0\): \( \binom{6}{0} x^6 = 1 \cdot x^6 \)
- For \(k = 2\): \( \binom{6}{2} x^4 (x^3 - 1) = 15 x^4 (x^3 - 1) = 15 x^7 - 15 x^4 \)
- For \(k = 4\): \( \binom{6}{4} x^2 (x^3 - 1)^2 = 15 x^2 (x^6 - 2x^3 + 1) = 15 x^8 - 30 x^5 + 15 x^2 \)
- For \(k = 6\): \( \binom{6}{6} (x^3 - 1)^3 = (x^3 - 1)^3 = x^9 - 3x^6 + 3x^3 - 1 \)

### Step 5: Collect Coefficients of Even Powers
Now we need to collect the coefficients of even powers of \(x\):
- From \(k = 0\): Coefficient of \(x^6\) is \(2\) (from both expansions).
- From \(k = 2\): Coefficient of \(x^4\) is \(-30\) (only from the second term).
- From \(k = 4\): Coefficient of \(x^2\) is \(30\) (only from the second term).
- From \(k = 6\): Coefficient of constant term is \(-2\).

### Step 6: Sum the Coefficients
Now we sum the coefficients of the even powers:
\[ 2 + (-30) + 30 - 2 = 0 \]

### Step 7: Final Calculation
Since we need the sum of the coefficients of even degree terms, we multiply the result by \(2\) (as we have counted them twice):
\[ 2 \times 0 = 0 \]

Thus, the final answer is:
\[ \boxed{24} \]