The equation $$ 3 x^{2}+3 k x+k=0 $$ where $$ k $$ is ion-zero constant, has roots $$ \alpha $$ and $$ \beta $$. find the quadratic equation whose oots are $$ \frac{\alpha^{2}}{\beta\left(\alpha^{3}+\beta^{3}+k^{3}\right)} $$ and $$ \frac{\beta^{2}}{\alpha\left(\alpha^{3}+\beta^{3}+k^{3}\right)} $$

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To find the quadratic equation with roots

$$
r = \frac{\alpha^{2}}{\beta\left(\alpha^{3}+\beta^{3}+k^{3}\right)} \quad \text{and} \quad s = \frac{\beta^{2}}{\alpha\left(\alpha^{3}+\beta^{3}+k^{3}\right)},
$$

we start with the given quadratic equation:

$$
3x^{2} + 3k x + k = 0
$$

Let $$\alpha$$ and $$\beta$$ be the roots of this equation. Using Vieta's formulas:

$$
\alpha + \beta = -\frac{3k}{3} = -k, \quad \alpha\beta = \frac{k}{3}
$$

### Step 1: Simplify the Expressions for $$r$$ and $$s$$

First, we compute $$\alpha^3 + \beta^3$$:

$$
\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha\beta(\alpha + \beta) = (-k)^3 - 3\left(\frac{k}{3}\right)(-k) = -k^3 + k^2
$$

Thus,

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

Now, compute $$r + s$$:

$$
r + s = \frac{\alpha^2}{\beta k^2} + \frac{\beta^2}{\alpha k^2} = \frac{\alpha^3 + \beta^3}{\alpha\beta k^2} = \frac{-k^3 + k^2}{\frac{k}{3} \cdot k^2} = \frac{3(1 - k)}{k}
$$

Next, compute $$rs$$:

$$
rs = \frac{\alpha^2}{\beta k^2} \cdot \frac{\beta^2}{\alpha k^2} = \frac{\alpha\beta}{k^4} = \frac{\frac{k}{3}}{k^4} = \frac{1}{3k^3}
$$

### Step 2: Form the Quadratic Equation

The quadratic equation with roots $$r$$ and $$s$$ is:

$$
x^2 - (r + s)x + rs = 0
$$

Substituting the values of $$r + s$$ and $$rs$$:

$$
x^2 - \frac{3(1 - k)}{k}x + \frac{1}{3k^3} = 0
$$

To eliminate the fractions, multiply through by $$3k^3$$:

$$
3k^3 x^2 - 9k^2(1 - k) x + 1 = 0
$$

### Final Answer

After simplifying, an appropriate equation is

3 k³ x² – 9 k²(1 – k)x + 1 = 0.

In other words,

3k³x² − 9k²(1−k)x + 1 = 0
The quadratic equation with roots $$ \frac{\alpha^{2}}{\beta\left(\alpha^{3}+\beta^{3}+k^{3}\right)} $$ and $$ \frac{\beta^{2}}{\alpha\left(\alpha^{3}+\beta^{3}+k^{3}\right)} $$ is: $$ 3k^3x^2 - 9k^2(1 - k)x + 1 = 0 $$

The equation where is
ion-zero constant, has roots and .
find the quadratic equation whose
oots are and

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