If ` 4 a - (4)/( a) = 3 ` then the value of : ` a^(3) - (1)/( a^(3) ) + 3 = ` ?

Options

A. `(3)/(16)`
B. `(7)/(16)`
C. `(21)/(64)`
D. `(21)/(16)`

Answer

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

### Step 1: Rearrange the equation
Start with the given equation:
\[ 4a - \frac{4}{a} = 3 \]
Multiply through by \( a \) to eliminate the fraction:
\[ 4a^2 - 4 = 3a \]
Rearranging gives:
\[ 4a^2 - 3a - 4 = 0 \]

### Step 2: Solve the quadratic equation
Now we can use the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 4, b = -3, c = -4 \):
\[ b^2 - 4ac = (-3)^2 - 4(4)(-4) = 9 + 64 = 73 \]
Thus, the solutions for \( a \) are:
\[ a = \frac{3 \pm \sqrt{73}}{8} \]

### Step 3: Calculate \( a^3 - \frac{1}{a^3} \)
To find \( a^3 - \frac{1}{a^3} \), we can use the identity:
\[ a^3 - \frac{1}{a^3} = \left( a - \frac{1}{a} \right) \left( a^2 + 1 + \frac{1}{a^2} \right) \]
First, we need to find \( a - \frac{1}{a} \):
From \( 4a - \frac{4}{a} = 3 \), we can express \( a - \frac{1}{a} \):
\[ 4\left(a - \frac{1}{a}\right) = 3 \implies a - \frac{1}{a} = \frac{3}{4} \]

Next, we find \( a^2 + \frac{1}{a^2} \):
Using the identity:
\[ a^2 + \frac{1}{a^2} = \left(a - \frac{1}{a}\right)^2 + 2 \]
Substituting \( a - \frac{1}{a} = \frac{3}{4} \):
\[ a^2 + \frac{1}{a^2} = \left(\frac{3}{4}\right)^2 + 2 = \frac{9}{16} + 2 = \frac{9}{16} + \frac{32}{16} = \frac{41}{16} \]

Now we can substitute back to find \( a^3 - \frac{1}{a^3} \):
\[ a^3 - \frac{1}{a^3} = \left(\frac{3}{4}\right) \left(\frac{41}{16} + 1\right) = \left(\frac{3}{4}\right) \left(\frac{41}{16} + \frac{16}{16}\right) = \left(\frac{3}{4}\right) \left(\frac{57}{16}\right) \]
Calculating this gives:
\[ a^3 - \frac{1}{a^3} = \frac{171}{64} \]

### Step 4: Add 3 to the result
Now we add 3 to \( a^3 - \frac{1}{a^3} \):
\[ a^3 - \frac{1}{a^3} + 3 = \frac{171}{64} + 3 = \frac{171}{64} + \frac{192}{64} = \frac{363}{64} \]

### Final Answer
Thus, the final value is:
\[ \boxed{\frac{363}{64}} \]