The expression `[x+(x^3-1)^(1/2)]^5+[x-(x^3-1)^(1/2)]^5` is a polynomial of degree

Options

A. 5
B. 6
C. 6
D. 8

Answer

To find the degree of the polynomial given by the expression \([x + (x^3 - 1)^{1/2}]^5 + [x - (x^3 - 1)^{1/2}]^5\), we will follow these steps:

### Step 1: Rewrite the Expression
The expression can be rewritten as:
\[ E = [x + (x^3 - 1)^{1/2}]^5 + [x - (x^3 - 1)^{1/2}]^5 \]

### Step 2: Apply the Binomial Theorem
Using the Binomial Theorem, we can expand both terms:
\[ E = \sum_{k=0}^{5} \binom{5}{k} x^{5-k} \left((x^3 - 1)^{1/2}\right)^k + \sum_{k=0}^{5} \binom{5}{k} x^{5-k} \left(- (x^3 - 1)^{1/2}\right)^k \]

### Step 3: Identify Terms that Cancel
Notice that the terms where \(k\) is odd will cancel out due to the negative sign in the second expansion:
- Terms with \(k = 1, 3, 5\) will cancel each other.

### Step 4: Collect Remaining Terms
The remaining terms will be those where \(k\) is even:
\[ E = 2 \left( \binom{5}{0} x^5 + \binom{5}{2} x^3 (x^3 - 1) + \binom{5}{4} x (x^3 - 1)^2 \right) \]

### Step 5: Calculate Each Remaining Term
1. For \(k = 0\):
\[ 2 \cdot \binom{5}{0} x^5 = 2x^5 \]
2. For \(k = 2\):
\[ 2 \cdot \binom{5}{2} x^3 (x^3 - 1) = 10x^3 (x^3 - 1) = 10x^6 - 10x^3 \]
3. For \(k = 4\):
\[ 2 \cdot \binom{5}{4} x (x^3 - 1)^2 = 10x (x^6 - 2x^3 + 1) = 10x^7 - 20x^4 + 10x \]

### Step 6: Combine All Terms
Combining all the terms, we have:
\[ E = 2x^5 + (10x^6 - 10x^3) + (10x^7 - 20x^4 + 10x) \]
This simplifies to:
\[ E = 10x^7 + 10x^6 - 20x^4 + 2x^5 - 10x^3 + 10x \]

### Step 7: Determine the Degree
The highest degree term in the polynomial is \(10x^7\), which indicates that the degree of the polynomial is:
\[ \text{Degree} = 7 \]

### Final Answer
Thus, the expression \([x + (x^3 - 1)^{1/2}]^5 + [x - (x^3 - 1)^{1/2}]^5\) is a polynomial of degree **7**.

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