The sum of the co-efficients of all odd degree terms in the expansion of `(x+sqrt(x^3-1))^5+(x-(sqrt(x^3-1))^5, (x gt 1)`

Options

A. 0
B. 1
C. 2
D. 4

Answer

To solve the problem, we need to find the sum of the coefficients of all odd degree terms in the expansion of \( (x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5 \).

### Step 1: Use the Binomial Theorem
We can apply the binomial theorem to expand both terms:
\[ (x + \sqrt{x^3 - 1})^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} (\sqrt{x^3 - 1})^k \]
\[ (x - \sqrt{x^3 - 1})^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} (-\sqrt{x^3 - 1})^k \]

### Step 2: Combine the Expansions
Adding the two expansions together:
\[ (x + \sqrt{x^3 - 1})^5 + (x - \sqrt{x^3 - 1})^5 = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} \left((\sqrt{x^3 - 1})^k + (-\sqrt{x^3 - 1})^k\right) \]

### Step 3: Simplify the Expression
Notice that the terms where \( k \) is odd will cancel out, while the terms where \( k \) is even will double:
\[ = 2 \sum_{k \text{ even}} \binom{5}{k} x^{5-k} (x^3 - 1)^{k/2} \]

### Step 4: Identify Even Values of k
The even values of \( k \) from 0 to 5 are \( k = 0, 2, 4 \):
- For \( k = 0 \): \( \binom{5}{0} x^5 (x^3 - 1)^0 = x^5 \)
- For \( k = 2 \): \( \binom{5}{2} x^3 (x^3 - 1)^1 = 10x^3(x^3 - 1) = 10x^6 - 10x^3 \)
- For \( k = 4 \): \( \binom{5}{4} x^1 (x^3 - 1)^2 = 5x((x^3 - 1)^2) = 5x(x^6 - 2x^3 + 1) = 5x^7 - 10x^4 + 5x \)

### Step 5: Combine All Terms
Now, we combine all the terms:
\[ = x^5 + (10x^6 - 10x^3) + (5x^7 - 10x^4 + 5x) \]
\[ = 5x^7 + 10x^6 - 10x^4 + x^5 - 10x^3 + 5x \]

### Step 6: Identify Odd Degree Terms
The odd degree terms in the final expression are:
- \( -10x^3 \) (coefficient = -10)
- \( 5x \) (coefficient = 5)

### Step 7: Calculate the Sum of Coefficients of Odd Degree Terms
Now we sum the coefficients of the odd degree terms:
\[ -10 + 5 = -5 \]

### Final Answer
Thus, the sum of the coefficients of all odd degree terms is:
\[ \boxed{-5} \]