Solve the equation $3x^3 - 11x^2 + 3x + 2 = 0$, expressing non-integer roots in surd form. Hence, solve the equation $x^3 + 11x^2 + 9x - 18 = 0$, expressing non-integer roots in surd form.

1. Use the Factor Theorem to find a rational root of the cubic equation.

Let $f(x) = 3x^3 - 11x^2 + 3x + 2$.
By trial and error with factors of the constant term over factors of the leading coefficient:
$f\left(\frac{2}{3}\right) = 3\left(\frac{8}{27}\right) - 11\left(\frac{4}{9}\right) + 3\left(\frac{2}{3}\right) + 2$
$= \frac{8}{9} - \frac{44}{9} + 2 + 2$
$= -\frac{36}{9} + 4 = -4 + 4 = 0$
Therefore, $x = \frac{2}{3}$ is a root, and $(3x - 2)$ is a factor.

2. Factorise the cubic polynomial into a linear and a quadratic factor.

$3x^3 - 11x^2 + 3x + 2 = (3x - 2)(ax^2 + bx + c)$
By comparing coefficients:
Leading term: $3a = 3 \implies a = 1$
Constant term: $-2c = 2 \implies c = -1$
$x^2$ term: $-2a + 3b = -11 \implies -2(1) + 3b = -11 \implies 3b = -9 \implies b = -3$
So, $3x^3 - 11x^2 + 3x + 2 = (3x - 2)(x^2 - 3x - 1) = 0$.

3. Solve the quadratic factor using the quadratic formula.

For $x^2 - 3x - 1 = 0$:
$x = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-1)}}{2(1)}$
$x = \frac{3 \pm \sqrt{9 + 4}}{2}$
$x = \frac{3 \pm \sqrt{13}}{2}$

4. State all the roots of the equation.

The roots are $x = \frac{2}{3}$ and $x = \frac{3 \pm \sqrt{13}}{2}$.

1. Identify the substitution that transforms the equation in (b) into the equation in (a).

Let $x = ky$. Substituting into $x^3 + 11x^2 + 9x - 18 = 0$ gives:
$k^3y^3 + 11k^2y^2 + 9ky - 18 = 0$
Dividing by $-9$ to match the constant term $+2$ in part (a):
$-\frac{k^3}{9}y^3 - \frac{11k^2}{9}y^2 - ky + 2 = 0$
Comparing the coefficient of $y$ with $3y^3 - 11y^2 + 3y + 2 = 0$, we need $-k = 3 \implies k = -3$.
Checking the other coefficients with $k = -3$:
$-\frac{(-3)^3}{9} = 3$ and $-\frac{11(-3)^2}{9} = -11$.
Thus, the substitution is $x = -3y$.

2. Use the roots from part (a) to find the roots of the new equation.

From part (a), the roots of $3y^3 - 11y^2 + 3y + 2 = 0$ are $y = \frac{2}{3}$ and $y = \frac{3 \pm \sqrt{13}}{2}$.
Since $x = -3y$, the roots of the new equation are:
$x = -3\left(\frac{2}{3}\right) = -2$
$x = -3\left(\frac{3 \pm \sqrt{13}}{2}\right) = \frac{-9 \pm 3\sqrt{13}}{2}$