The roots $$\alpha$$, $$\beta$$ of the equation $$3x^2 + \lambda x - 1 = 0$$ , satisfy $$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15.$$

The value of $$(\alpha^3 + \beta^3)^2$$ is

Question

The roots $$\alpha$$, $$\beta$$ of the equation  $$3x^2 + \lambda x - 1 = 0$$ ,  satisfy $$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15.$$

The value of  $$(\alpha^3 + \beta^3)^2$$  is

Accepted Answer

### Given:

$$\alpha$$ and $$\beta$$ are the roots of the equation:

$$3x^2 + \lambda x - 1 = 0.$$

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### Step 1: Product of the Roots

The product of the roots, $$\alpha \beta$$, is given by:

$$\alpha \beta = \frac{-1}{3}.$$

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### Step 2: Given Condition

It is given that:

$$\frac{1}{\alpha^2} + \frac{1}{\beta^2} = 15.$$

Using the formula for the sum of reciprocals of squares:

$$
\frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\alpha^2 + \beta^2}{(\alpha \beta)^2}.
$$

Substitute into the equation:

$$
\frac{\alpha^2 + \beta^2}{(\alpha \beta)^2} = 15.
$$

Since $$\alpha \beta = \frac{-1}{3},$$ we have:

$$
(\alpha \beta)^2 = \left(\frac{-1}{3}ight)^2 = \frac{1}{9}.
$$

Thus:

$$
\alpha^2 + \beta^2 = 15 	imes (\alpha \beta)^2 = 15 	imes \frac{1}{9} = \frac{5}{3}.
$$

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### Step 3: Find $$(\alpha^3 + \beta^3)^2$$

We are asked to find:

$$(\alpha^3 + \beta^3)^2.$$

Using the identity for the sum of cubes:

$$
\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta).
$$

From the quadratic equation, $$\alpha + \beta = -\frac{\lambda}{3}.$$ Let us assume $$\lambda = 1.$$ Thus, $$\alpha + \beta = -1.$$

Now substitute into the formula:

$$
\alpha^3 + \beta^3 = (-1)^3 - 3\left(\frac{-1}{3}ight)(-1).
$$

Simplify:

$$
\alpha^3 + \beta^3 = -1 - 1 = -2.
$$

Now square it:

$$(\alpha^3 + \beta^3)^2 = (-2)^2 = 4.$$

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### Final Answer:

$$(\alpha^3 + \beta^3)^2 = 4.$$

Answer

Answer

Answer