Here's the solution to the problem: 1. (i) a) Given that the roots of the equation $3x^2 - 4x + 2 = 0$ are $\alpha$ and $\beta$. Show that $\alpha^3 + \beta^3 = -\frac{8}{27}$. Step 1: Identify the coefficients and find the sum and product of roots. 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: $$ \alpha + \beta = -\frac{-4}{3} = \frac{4}{3} $$ Product of roots: $$ \alpha \beta = \frac{2}{3} $$ Step 2: Use the identity for $\alpha^3 + \beta^3$. 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: $$ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 2\alpha \beta - \alpha \beta) $$ $$ \alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta) $$ Step 3: Substitute the values of $\alpha + \beta$ and $\alpha \beta$. $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\left(\frac{4}{3}\right)^2 - 3\left(\frac{2}{3}\right)\right) $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\frac{16}{9} - \frac{6}{3}\right) $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\frac{16}{9} - \frac{18}{9}\right) $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(-\frac{2}{9}\right) $$ $$ \alpha^3 + \beta^3 = -\frac{8}{27} $$ This shows the required result. 1. (i) b) Hence, find the equation with integral coefficients whose roots are $\frac{1}{\alpha^3}$ and $\frac{1}{\beta^3}$. Step 1: Define the new roots and calculate their sum. Let the new roots be $p = \frac{1}{\alpha^3}$ and $q = \frac{1}{\beta^3}$. Sum of new roots: $$ p + q = \frac{1}{\alpha^3} + \frac{1}{\beta^3} = \frac{\beta^3 + \alpha^3}{\alpha^3 \beta^3} $$ From part (a), we know $\alpha^3 + \beta^3 = -\frac{8}{27}$. Also, $\alpha^3 \beta^3 = (\alpha \beta)^3$. Since $\alpha \beta = \frac{2}{3}$: $$ (\alpha \beta)^3 = \left(\frac{2}{3}\right)^3 = \frac{8}{27} $$ Now, substitute these values into the sum of new roots: $$ p + q = \frac{-\frac{8}{27}}{\frac{8}{27}} = -1 $$ Step 2: Calculate the product of the new roots. $$ pq = \left(\frac{1}{\alpha^3}\right)\left(\frac{1}{\beta^3}\right) = \frac{1}{\alpha^3 \beta^3} = \frac{1}{(\alpha \beta)^3} $$ $$ pq = \frac{1}{\frac{8}{27}} = \frac{27}{8} $$ Step 3: Form the quadratic equation with integral coefficients. The general form of a quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$. $$ x^2 - (-1)x + \frac{27}{8} = 0 $$ $$ x^2 + x + \frac{27}{8} = 0 $$ To obtain integral coefficients, multiply the entire equation by 8: $$ 8(x^2 + x + \frac{27}{8}) = 8(0) $$ $$ \boxed{8x^2 + 8x + 27 = 0} $$ 1. (iii) Find the set of real values of $x$ for which $\left|\frac{x+3}{x-5}\right| = \frac{x+3}{x-5}$. Step 1: Understand the property of absolute value. The equation $|A| = A$ holds true if and only if $A \ge 0$. Therefore, we need to find the values of $x$ for which $\frac{x+3}{x-5} \ge 0$. Step 2: Find the critical points. 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 $(5, \infty)$. Step 3: Test each interval. • 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. Step 4: Combine the valid intervals. The set of real values of $x$ for which the inequality holds is $x \le -3$ or $x > 5$. In set notation: $$ \boxed{\{x \in \mathbb{R} \mid x \le -3 \text{ or } x > 5\}} $$