Consider biquadratic equation `81x^(4) + 216x^(3) + 96x = 65`, whose roots are `a,b,c,d`. Given `a,b`, real roots and `c,d` are imaginary roots.
On the basis of above information, answer the followin questions:
The Value of `c^(3) + d^(3) - (a+b))^(3)` is equal to

Options

A. `(52)/(9)`
B. `(53)/(9)`
C. `(59)/(9)`
D. `(50)/(9)`

Explanation

To solve the given biquadratic equation \( 81x^{4} + 216x^{3} + 96x = 65 \) and find the value of \( c^{3} + d^{3} - (a + b)^{3} \), we will follow these steps:

### Step 1: Rewrite the Equation
First, we rewrite the equation in standard form:
\[ 81x^{4} + 216x^{3} + 96x - 65 = 0 \]

### Step 2: Factor the Equation
Next, we will factor the polynomial. We can rewrite it as:
\[ 81x^{4} + 216x^{3} + 96x - 65 = 0 \]
Factoring gives us:
\[ (3x + 2)^{4} = 81 \]

### Step 3: Set Up for Solving
We can rewrite \( 81 \) as \( 9^{2} \):
\[ (3x + 2)^{4} - 9^{2} = 0 \]
This can be factored using the difference of squares:
\[ ((3x + 2)^{2} - 9)((3x + 2)^{2} + 9) = 0 \]

### Step 4: Solve for Real Roots
From the first factor:
\[ (3x + 2)^{2} - 9 = 0 \implies (3x + 2)^{2} = 9 \]
Taking the square root gives:
\[ 3x + 2 = 3 \quad \text{or} \quad 3x + 2 = -3 \]
Solving these:
1. \( 3x + 2 = 3 \) leads to \( x = \frac{1}{3} \) (real root \( a \)).
2. \( 3x + 2 = -3 \) leads to \( x = -\frac{5}{3} \) (real root \( b \)).

### Step 5: Solve for Imaginary Roots
From the second factor:
\[ (3x + 2)^{2} + 9 = 0 \implies (3x + 2)^{2} = -9 \]
Taking the square root gives:
\[ 3x + 2 = \pm 3i \]
Solving these:
1. \( 3x + 2 = 3i \) leads to \( x = -\frac{2}{3} + i \) (imaginary root \( c \)).
2. \( 3x + 2 = -3i \) leads to \( x = -\frac{2}{3} - i \) (imaginary root \( d \)).

### Step 6: Calculate \( c^{3} + d^{3} \)
Using the identity for cubes:
\[ c^{3} + d^{3} = (c + d)(c^{2} - cd + d^{2}) \]
We find:
- \( c + d = \left(-\frac{2}{3} + i\right) + \left(-\frac{2}{3} - i\right) = -\frac{4}{3} \)
- \( cd = \left(-\frac{2}{3}\right)^{2} + 1 = \frac{4}{9} + 1 = \frac{13}{9} \)

Now calculate \( c^{2} + d^{2} \):
\[ c^{2} + d^{2} = (c + d)^{2} - 2cd = \left(-\frac{4}{3}\right)^{2} - 2 \cdot \frac{13}{9} = \frac{16}{9} - \frac{26}{9} = -\frac{10}{9} \]

Thus:
\[ c^{3} + d^{3} = \left(-\frac{4}{3}\right)\left(-\frac{10}{9} - \frac{13}{9}\right) = -\frac{4}{3} \cdot -\frac{23}{9} = \frac{92}{27} \]

### Step 7: Calculate \( (a + b)^{3} \)
Now calculate \( a + b \):
\[ a + b = \frac{1}{3} - \frac{5}{3} = -\frac{4}{3} \]
Then:
\[ (a + b)^{3} = \left(-\frac{4}{3}\right)^{3} = -\frac{64}{27} \]

### Step 8: Final Calculation
Now we can find:
\[ c^{3} + d^{3} - (a + b)^{3} = \frac{92}{27} + \frac{64}{27} = \frac{156}{27} = \frac{52}{9} \]

### Conclusion
Thus, the value of \( c^{3} + d^{3} - (a + b)^{3} \) is:
\[ \boxed{\frac{52}{9}} \]