Given equation: 3x² − 4x + 2 = 0.

For a quadratic equation $ax^2 + bx + c = 0$, the sum of roots is $\alpha + \beta = -\frac{b}{a}$ and the product of roots is $\alpha \beta = \frac{c}{a}$.
Given equation: $3x^2 - 4x + 2 = 0$.
Here, $a=3$, $b=-4$, $c=2$.
Sum of roots:

Product of roots:

The identity for the sum of cubes is $\alpha^3 + \beta^3 = (\alpha + \beta)(\alpha^2 - \alpha \beta + \beta^2)$.
We can express $\alpha^2 + \beta^2$ in terms of $\alpha + \beta$ and $\alpha \beta$: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.
Substitute this into the identity:

This shows the required result.

1. (i) b) Hence, find the equation with integral coefficients whose roots are $\frac{1}{\alpha^3}$ and $\frac{1}{\beta^3}$.

Let the new roots be $p = \frac{1}{\alpha^3}$ and $q = \frac{1}{\beta^3}$.
Sum of new roots:

From part (a), we know $\alpha^3 + \beta^3 = -\frac{8}{27}$.
Also, $\alpha^3 \beta^3 = (\alpha \beta)^3$. Since $\alpha \beta = \frac{2}{3}$:

Now, substitute these values into the sum of new roots:

The general form of a quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$.

To obtain integral coefficients, multiply the entire equation by 8:

1. (iii) Find the set of real values of $x$ for which $\left|\frac{x+3}{x-5}\right| = \frac{x+3}{x-5}$.

The equation $|A| = A$ holds true if and only if $A \ge 0$.
Therefore, we need to find the values of $x$ for which $\frac{x+3}{x-5} \ge 0$.

The critical points are the values of $x$ where the numerator or denominator is zero.
Numerator: $x+3 = 0 \implies x = -3$.
Denominator: $x-5 = 0 \implies x = 5$.
These points divide the number line into three intervals: $(-\infty, -3)$, $(-3, 5)$, and $(5, \infty)$.

• For $x < -3$ (e.g., $x = -4$): $\frac{-4+3}{-4-5} = \frac{-1}{-9} = \frac{1}{9}$. This is positive, so the inequality holds. Since the inequality is $\ge 0$, $x=-3$ is included (as $\frac{0}{-8}=0$). So, $x \le -3$ is part of the solution.

• For $-3 < x < 5$ (e.g., $x = 0$): $\frac{0+3}{0-5} = \frac{3}{-5} = -\frac{3}{5}$. This is negative, so the inequality does not hold.

• For $x > 5$ (e.g., $x = 6$): $\frac{6+3}{6-5} = \frac{9}{1} = 9$. This is positive, so the inequality holds. Note that $x=5$ is excluded because it makes the denominator zero, which is undefined.

The set of real values of $x$ for which the inequality holds is $x \le -3$ or $x > 5$.
In set notation: