Here's the solution to question 1: 1(i)a) Show that $\alpha^3 + \beta^3 = -\frac{8}{27}$ Given the quadratic equation $3x^2 - 4x + 2 = 0$, with roots $\alpha$ and $\beta$. 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 $\alpha \beta = \frac{c}{a}$. Here, $a=3$, $b=-4$, $c=2$. $$ \alpha + \beta = -\frac{-4}{3} = \frac{4}{3} $$ $$ \alpha \beta = \frac{2}{3} $$ Step 2: Use the identity for $\alpha^3 + \beta^3$. The identity is $\alpha^3 + \beta^3 = (\alpha + \beta)((\alpha + \beta)^2 - 3\alpha \beta)$. Step 3: Substitute the values of $\alpha + \beta$ and $\alpha \beta$. $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\left(\frac{4}{3}\right)^2 - 3\left(\frac{2}{3}\right)\right) $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\frac{16}{9} - 2\right) $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(\frac{16}{9} - \frac{18}{9}\right) $$ $$ \alpha^3 + \beta^3 = \left(\frac{4}{3}\right)\left(-\frac{2}{9}\right) $$ $$ \alpha^3 + \beta^3 = -\frac{8}{27} $$ This shows the required result. 1(i)b) Find the equation with integral coefficients whose roots are $\frac{1}{\alpha^2}$ and $\frac{1}{\beta^2}$ Let the new roots be $r_1 = \frac{1}{\alpha^2}$ and $r_2 = \frac{1}{\beta^2}$. The general quadratic equation is $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$. Step 1: Calculate the sum of the new roots. $$ S' = \frac{1}{\alpha^2} + \frac{1}{\beta^2} = \frac{\beta^2 + \alpha^2}{(\alpha \beta)^2} $$ First, find $\alpha^2 + \beta^2$: $$ \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta $$ Substitute the values from part (a): $$ \alpha^2 + \beta^2 = \left(\frac{4}{3}\right)^2 - 2\left(\frac{2}{3}\right) = \frac{16}{9} - \frac{4}{3} = \frac{16}{9} - \frac{12}{9} = \frac{4}{9} $$ Now, calculate $S'$: $$ S' = \frac{\frac{4}{9}}{\left(\frac{2}{3}\right)^2} = \frac{\frac{4}{9}}{\frac{4}{9}} = 1 $$ Step 2: Calculate the product of the new roots. $$ P' = \left(\frac{1}{\alpha^2}\right)\left(\frac{1}{\beta^2}\right) = \frac{1}{(\alpha \beta)^2} $$ Substitute the value of $\alpha \beta$: $$ P' = \frac{1}{\left(\frac{2}{3}\right)^2} = \frac{1}{\frac{4}{9}} = \frac{9}{4} $$ Step 3: Form the new quadratic equation. The equation is $x^2 - S'x + P' = 0$. $$ x^2 - 1x + \frac{9}{4} = 0 $$ To obtain integral coefficients, multiply the entire equation by 4: $$ 4x^2 - 4x + 9 = 0 $$ The equation with integral coefficients is $\boxed{\text{4x^2 - 4x + 9 = 0}}$. 1(ii) Find the set of real values of $x$ for which $\left|\frac{x+3}{x-1}\right| = \frac{x+3}{x-1}$ The property $|A| = A$ holds true if and only if $A \ge 0$. Therefore, we need to find the values of $x$ for which $\frac{x+3}{x-1} \ge 0$. Also, the denominator cannot be zero, so $x-1 \ne 0 \implies x \ne 1$. Step 1: Analyze the signs of the numerator and denominator. For the fraction to be non-negative, both numerator and denominator must have the same sign (or the numerator is zero). Case 1: Numerator is non-negative and denominator is positive. $x+3 \ge 0 \implies x \ge -3$ $x-1 > 0 \implies x > 1$ The intersection of these conditions is $x > 1$. Case 2: Numerator is non-positive and denominator is negative. $x+3 \le 0 \implies x \le -3$ $x-1 < 0 \implies x < 1$ The intersection of these conditions is $x \le -3$. Step 2: Combine the solutions from both cases. The set of real values of $x$ is $x \le -3$ or $x > 1$. In interval notation, this is $(-\infty, -3] \cup (1, \infty)$. The set of real values of $x$ is $\boxed{x \le -3 \text{ or } x > 1}$.