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

Explanation

To determine 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
We can rewrite the expression using \( a = x \) and \( b = (x^3 - 1)^{1/2} \):

\[ (a + b)^5 + (a - b)^5. \]

### Step 2: Apply the Binomial Theorem
Using the binomial theorem, we expand both terms:

1. For \( (a + b)^5 \):
\[ (a + b)^5 = \sum_{k=0}^{5} \binom{5}{k} a^{5-k} b^k. \]

2. For \( (a - b)^5 \):
\[ (a - b)^5 = \sum_{k=0}^{5} \binom{5}{k} a^{5-k} (-b)^k = \sum_{k=0}^{5} \binom{5}{k} a^{5-k} (-1)^k b^k. \]

### Step 3: Combine the Expansions
Now, we combine both expansions:

\[ (a + b)^5 + (a - b)^5 = \sum_{k=0}^{5} \binom{5}{k} a^{5-k} b^k + \sum_{k=0}^{5} \binom{5}{k} a^{5-k} (-1)^k b^k. \]

### Step 4: Simplify the Expression
Notice that terms where \( k \) is odd will cancel out, while terms where \( k \) is even will double:

\[ = 2 \sum_{k \text{ even}} \binom{5}{k} a^{5-k} b^k. \]

The even values of \( k \) are \( 0, 2, 4 \).

### Step 5: Identify the Highest Degree Term
Now, we calculate the highest degree terms:

- For \( k = 0 \):
\[ 2 \binom{5}{0} a^5 b^0 = 2x^5. \]

- For \( k = 2 \):
\[ 2 \binom{5}{2} a^3 b^2 = 2 \cdot 10 x^3 (x^3 - 1) = 20 x^3 (x^3 - 1) = 20 x^6 - 20 x^3. \]

- For \( k = 4 \):
\[ 2 \binom{5}{4} a^1 b^4 = 2 \cdot 5 x (x^3 - 1)^2. \]

The term \( (x^3 - 1)^2 \) expands to \( x^6 - 2x^3 + 1 \), leading to:

\[ 10 x^7 - 20 x^4 + 10 x. \]

### Step 6: Combine All Terms
Now, we combine all the highest degree terms:

- From \( k = 0 \): \( 2x^5 \)
- From \( k = 2 \): \( 20x^6 - 20x^3 \)
- From \( k = 4 \): \( 10x^7 - 20x^4 + 10x \)

The highest degree term is \( 10x^7 \).

### Conclusion
Thus, the degree of the polynomial is:

\[ \text{Degree} = 7. \]