If the equation `4y^(3) - 8a^(2)yx^(2) - 3ay^(2)x +8x^(3) = 0` represents three straight lines, two of them are perpendicular, then sum of all possible values of a is equal to

Options

A. `(3)/(8)`
B. `(-3)/(4)`
C. `(1)/(4)`
D. `-2`

Explanation

To solve the problem, we need to analyze the given cubic equation in terms of the variable \( y \) and \( x \):

Given equation:
\[ 4y^3 - 8a^2yx^2 - 3ay^2x + 8x^3 = 0 \]

This represents three straight lines in the \( y \)-\( x \) plane. For the lines to be perpendicular, we will use the condition that the product of the slopes of the two perpendicular lines is \(-1\).

### Step 1: Rewrite the equation
We can rewrite the equation in the standard form of a cubic polynomial in \( y \):
\[ 4y^3 - 3ay^2x - 8a^2yx^2 + 8x^3 = 0 \]

### Step 2: Identify coefficients
Let \( m \) be the slope of the lines. The equation can be expressed as:
\[ 4m^3 - 3am^2 - 8a^2m + 8 = 0 \]
where \( m \) represents the slopes of the lines.

### Step 3: Use Vieta's formulas
According to Vieta's formulas, for a cubic equation \( Am^3 + Bm^2 + Cm + D = 0 \):
- The sum of the roots \( m_1 + m_2 + m_3 = -\frac{B}{A} \)
- The sum of the product of the roots taken two at a time \( m_1m_2 + m_2m_3 + m_3m_1 = \frac{C}{A} \)
- The product of the roots \( m_1m_2m_3 = -\frac{D}{A} \)

For our equation:
- \( A = 4 \)
- \( B = -3a \)
- \( C = -8a^2 \)
- \( D = 8 \)

### Step 4: Calculate the sums and products
From Vieta's:
1. \( m_1 + m_2 + m_3 = \frac{3a}{4} \)
2. \( m_1m_2 + m_2m_3 + m_3m_1 = -\frac{8a^2}{4} = -2a^2 \)
3. \( m_1m_2m_3 = -\frac{8}{4} = -2 \)

### Step 5: Condition for perpendicular lines
If two lines are perpendicular, say \( m_1 \) and \( m_2 \), then:
\[ m_1m_2 = -1 \]

### Step 6: Substitute and solve
Using the product of the roots:
\[ m_1m_2m_3 = -2 \implies m_3 = \frac{-2}{m_1m_2} = \frac{-2}{-1} = 2 \]

Now substituting \( m_3 = 2 \) into the sum of the roots:
\[ m_1 + m_2 + 2 = \frac{3a}{4} \implies m_1 + m_2 = \frac{3a}{4} - 2 \]

Using the sum of the product of the roots:
\[ m_1m_2 + 2(m_1 + m_2) = -2a^2 \]
Substituting \( m_1m_2 = -1 \):
\[ -1 + 2\left(\frac{3a}{4} - 2\right) = -2a^2 \]
Expanding and simplifying:
\[ -1 + \frac{3a}{2} - 4 = -2a^2 \implies \frac{3a}{2} - 5 = -2a^2 \]
Rearranging gives:
\[ 2a^2 + \frac{3a}{2} + 5 = 0 \]
Multiplying through by 2 to eliminate the fraction:
\[ 4a^2 + 3a + 10 = 0 \]

### Step 7: Solve the quadratic equation
Using the quadratic formula:
\[ a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 4 \cdot 10}}{2 \cdot 4} \]
Calculating the discriminant:
\[ 9 - 160 = -151 \]
Since the discriminant is negative, there are no real roots for \( a \).

### Final Answer
Thus, the sum of all possible values of \( a \) is \( 0 \) since there are no real solutions.