If `x = 2 + sqrt(5)` then the value of `(x^(3) - x^(-3))` is :

Options

A. `-52`
B. 52
C. `46sqrt(5)`
D. `-76`

Answer

To solve the problem, we need to find the value of \( x^3 - x^{-3} \) given that \( x = 2 + \sqrt{5} \).

### Step 1: Calculate \( x^3 \)

First, we need to find \( x^3 \). We can use the binomial expansion or directly calculate it.

\[ x = 2 + \sqrt{5} \]

Now, we calculate \( x^2 \) first:

\[ x^2 = (2 + \sqrt{5})^2 = 2^2 + 2 \cdot 2 \cdot \sqrt{5} + (\sqrt{5})^2 = 4 + 4\sqrt{5} + 5 = 9 + 4\sqrt{5} \]

Now, we can find \( x^3 \):

\[ x^3 = x \cdot x^2 = (2 + \sqrt{5})(9 + 4\sqrt{5}) \]

Using the distributive property:

\[ x^3 = 2 \cdot 9 + 2 \cdot 4\sqrt{5} + \sqrt{5} \cdot 9 + \sqrt{5} \cdot 4\sqrt{5} \]
\[ = 18 + 8\sqrt{5} + 9\sqrt{5} + 4 \cdot 5 \]
\[ = 18 + 8\sqrt{5} + 9\sqrt{5} + 20 \]
\[ = 38 + 17\sqrt{5} \]

### Step 2: Calculate \( x^{-1} \)

Next, we find \( x^{-1} \):

\[ x^{-1} = \frac{1}{x} = \frac{1}{2 + \sqrt{5}} \]

To rationalize the denominator, multiply the numerator and denominator by \( 2 - \sqrt{5} \):

\[ x^{-1} = \frac{2 - \sqrt{5}}{(2 + \sqrt{5})(2 - \sqrt{5})} = \frac{2 - \sqrt{5}}{4 - 5} = 2 - \sqrt{5} \]

### Step 3: Calculate \( x^{-3} \)

Now we calculate \( x^{-3} \):

\[ x^{-3} = (x^{-1})^3 = (2 - \sqrt{5})^3 \]

Using the binomial expansion:

\[ (2 - \sqrt{5})^3 = 2^3 - 3 \cdot 2^2 \cdot \sqrt{5} + 3 \cdot 2 \cdot (\sqrt{5})^2 - (\sqrt{5})^3 \]
\[ = 8 - 12\sqrt{5} + 30 - 5\sqrt{5} \]
\[ = 38 - 17\sqrt{5} \]

### Step 4: Calculate \( x^3 - x^{-3} \)

Now we can find \( x^3 - x^{-3} \):

\[ x^3 - x^{-3} = (38 + 17\sqrt{5}) - (38 - 17\sqrt{5}) \]
\[ = 38 + 17\sqrt{5} - 38 + 17\sqrt{5} \]
\[ = 34\sqrt{5} \]

### Final Answer

Thus, the value of \( x^3 - x^{-3} \) is:

\[ \boxed{34\sqrt{5}} \]