If ` 4 b^(2) + (1)/( b^(2)) = 2 ` then the value of ` 8 b^(3) + (1)/( b^(3))` is

Options

A. 0
B. 1
C. 2
D. 5

Answer

To solve the equation \( 4b^2 + \frac{1}{b^2} = 2 \) and find the value of \( 8b^3 + \frac{1}{b^3} \), we can follow these steps:

### Step 1: Rearrange the given equation
We start with the equation:
\[ 4b^2 + \frac{1}{b^2} = 2 \]
To make it easier to work with, we can add 4 to both sides:
\[ 4b^2 + \frac{1}{b^2} + 4 = 6 \]

### Step 2: Recognize the perfect square
Notice that \( 4b^2 + \frac{1}{b^2} + 4 \) can be expressed as a perfect square:
\[ (2b + \frac{1}{b})^2 = 6 \]
This means:
\[ 2b + \frac{1}{b} = \sqrt{6} \quad \text{or} \quad 2b + \frac{1}{b} = -\sqrt{6} \]
Since \( b \) is a real number, we will consider only the positive root:
\[ 2b + \frac{1}{b} = \sqrt{6} \]

### Step 3: Cube the expression
Now, we want to find \( 8b^3 + \frac{1}{b^3} \). We can use the identity for the cube of a binomial:
\[ (a + b)^3 = a^3 + b^3 + 3ab(a + b) \]
Let \( a = 2b \) and \( b = \frac{1}{b} \):
\[ (2b + \frac{1}{b})^3 = (2b)^3 + \left(\frac{1}{b}\right)^3 + 3(2b)\left(\frac{1}{b}\right)(2b + \frac{1}{b}) \]
This simplifies to:
\[ (2b + \frac{1}{b})^3 = 8b^3 + \frac{1}{b^3} + 3(2)(1)(2b + \frac{1}{b}) \]
Substituting \( 2b + \frac{1}{b} = \sqrt{6} \):
\[ (\sqrt{6})^3 = 8b^3 + \frac{1}{b^3} + 3(2)(\sqrt{6}) \]

### Step 4: Calculate \( \sqrt{6}^3 \)
Calculating \( \sqrt{6}^3 \):
\[ \sqrt{6}^3 = 6\sqrt{6} \]

### Step 5: Substitute and simplify
Now we have:
\[ 6\sqrt{6} = 8b^3 + \frac{1}{b^3} + 6\sqrt{6} \]
Subtract \( 6\sqrt{6} \) from both sides:
\[ 8b^3 + \frac{1}{b^3} = 0 \]

### Final Answer
Thus, the value of \( 8b^3 + \frac{1}{b^3} \) is:
\[ \boxed{0} \]