If `sqrt(x) + (1)/(sqrt(x)) = sqrt(7)`, then `x^(3) + (1)/(x^(3))` is equal to :

Options

A. 140
B. 130
C. 120
D. 110

Answer

To solve the equation \( \sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{7} \) and find \( x^3 + \frac{1}{x^3} \), we can follow these steps:

### Step 1: Square both sides
Start with the given equation:
\[ \sqrt{x} + \frac{1}{\sqrt{x}} = \sqrt{7} \]
Now, square both sides:
\[ \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right)^2 = (\sqrt{7})^2 \]
This expands to:
\[ x + 2 + \frac{1}{x} = 7 \]

### Step 2: Rearrange the equation
Rearranging gives us:
\[ x + \frac{1}{x} = 7 - 2 = 5 \]

### Step 3: Use the identity for cubes
We know the identity:
\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \]
In our case, let \( a = \sqrt{x} \) and \( b = \frac{1}{\sqrt{x}} \). We can express \( x^3 + \frac{1}{x^3} \) in terms of \( x + \frac{1}{x} \):
\[ x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 - x \cdot \frac{1}{x} + \frac{1}{x^2}\right) \]
This simplifies to:
\[ x^3 + \frac{1}{x^3} = \left(x + \frac{1}{x}\right)\left(x^2 - 1 + \frac{1}{x^2}\right) \]

### Step 4: Find \( x^2 + \frac{1}{x^2} \)
To find \( x^2 + \frac{1}{x^2} \), we can use the square of \( x + \frac{1}{x} \):
\[ \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2} \]
Substituting \( x + \frac{1}{x} = 5 \):
\[ 5^2 = x^2 + 2 + \frac{1}{x^2} \]
This gives:
\[ 25 = x^2 + 2 + \frac{1}{x^2} \]
Thus,
\[ x^2 + \frac{1}{x^2} = 25 - 2 = 23 \]

### Step 5: Substitute back to find \( x^3 + \frac{1}{x^3} \)
Now, substitute \( x + \frac{1}{x} = 5 \) and \( x^2 + \frac{1}{x^2} = 23 \) into the identity:
\[ x^3 + \frac{1}{x^3} = (x + \frac{1}{x})\left(x^2 + \frac{1}{x^2} - 1\right) \]
This becomes:
\[ x^3 + \frac{1}{x^3} = 5(23 - 1) = 5 \times 22 = 110 \]

### Final Answer
Thus, \( x^3 + \frac{1}{x^3} = 110 \).