If sum of square of reciprocal of root `alpha` and `beta` of equation `3x^2+ lamda x-1=0` is `15` then find `6(alpha^3+beta^3)^2`

Options

A. 18
B. 24
C. 36
D. 96

Answer

To solve the problem step by step, we will follow the given information and apply relevant mathematical identities.

### Step 1: Understand the given information
We are given a quadratic equation:
\[ 3x^2 + \lambda x - 1 = 0 \]
Let the roots of this equation be \( \alpha \) and \( \beta \).

We know that:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15 \]

### Step 2: Use the identity for the sum of squares of reciprocals
The sum of squares of the reciprocals of the roots can be expressed as:
\[ \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{\alpha^2 \beta^2} \]

Using the identity \( \beta^2 + \alpha^2 = (\alpha + \beta)^2 - 2\alpha\beta \), we can rewrite the equation as:
\[ \frac{(\alpha + \beta)^2 - 2\alpha\beta}{(\alpha\beta)^2} = 15 \]

### Step 3: Apply Vieta's formulas
From Vieta's formulas for the quadratic equation \( 3x^2 + \lambda x - 1 = 0 \):
- The sum of the roots \( \alpha + \beta = -\frac{\lambda}{3} \)
- The product of the roots \( \alpha\beta = -\frac{1}{3} \)

### Step 4: Substitute the values into the equation
Substituting \( \alpha + \beta \) and \( \alpha \beta \) into the equation gives:
\[ \frac{\left(-\frac{\lambda}{3}\right)^2 - 2\left(-\frac{1}{3}\right)}{\left(-\frac{1}{3}\right)^2} = 15 \]

Calculating the denominator:
\[ \left(-\frac{1}{3}\right)^2 = \frac{1}{9} \]

Now substituting this into the equation:
\[ \frac{\frac{\lambda^2}{9} + \frac{2}{3}}{\frac{1}{9}} = 15 \]

### Step 5: Simplify the equation
Multiplying both sides by \( \frac{1}{9} \):
\[ \lambda^2 + 6 = 15 \]

### Step 6: Solve for \( \lambda \)
Rearranging gives:
\[ \lambda^2 = 9 \]
Thus, \( \lambda = 3 \) or \( \lambda = -3 \).

### Step 7: Calculate \( \alpha^3 + \beta^3 \)
Using the identity:
\[ \alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha\beta + \beta^2) \]
We already know:
- \( \alpha + \beta = -\frac{\lambda}{3} = -1 \) (for \( \lambda = 3 \))
- \( \alpha^2 + \beta^2 = 15 + 2\left(-\frac{1}{3}\right) = 15 + \frac{2}{3} = \frac{47}{3} \)

Now substituting:
\[ \alpha^3 + \beta^3 = -1 \left(\frac{47}{3} + \frac{1}{3}\right) = -1 \left(\frac{48}{3}\right) = -16 \]

### Step 8: Calculate \( 6(\alpha^3 + \beta^3)^2 \)
Now we need to find:
\[ 6(\alpha^3 + \beta^3)^2 = 6(-16)^2 = 6 \times 256 = 1536 \]

### Final Answer
Thus, the final answer is:
\[ \boxed{1536} \]