The diagram below shows the curve $y = x^3 - 12x^2 + 20x$, which crosses the $x$-axis at the origin $O$ and the points $A$ and $B$. The tangent to the curve at the point $(8, -96)$ crosses the $x$-axis at the point $C$.

1. Set $y = 0$ to find the $x$-intercepts and factorise the cubic expression

$x^3 - 12x^2 + 20x = 0$
$x(x^2 - 12x + 20) = 0$

2. Factorise the quadratic factor and solve for $x$

$x(x - 2)(x - 10) = 0$
$x = 0 \text{ or } x = 2 \text{ or } x = 10$

3. Identify the coordinates of $A$ and $B$ from the graph

From the graph, $A$ and $B$ are the positive $x$-intercepts, with $A$ being closer to the origin.
Therefore, $A$ is $(2, 0)$ and $B$ is $(10, 0)$.

1. Differentiate the equation of the curve to find the gradient function

$y = x^3 - 12x^2 + 20x$
$\frac{dy}{dx} = 3x^2 - 24x + 20$

2. Find the gradient of the tangent at $x = 8$

At $x = 8$,
$\frac{dy}{dx} = 3(8)^2 - 24(8) + 20$
$= 192 - 192 + 20 = 20$

3. Find the equation of the tangent

Using $y - y_1 = m(x - x_1)$ with point $(8, -96)$ and $m = 20$:
$y - (-96) = 20(x - 8)$
$y + 96 = 20x - 160$
$y = 20x - 256$

4. Find the coordinates of $C$ by setting $y = 0$

At $C$, the tangent crosses the $x$-axis, so $y = 0$.
$0 = 20x - 256$
$20x = 256$
$x = 12.8$
Coordinates of $C$ are $(12.8, 0)$.

1. Find the coordinates of $A$, $B$, and $C$ from earlier parts

From (i), $x^3 - 12x^2 + 20x = 0 \implies x(x-2)(x-10) = 0$, so $A$ is $(2, 0)$ and $B$ is $(10, 0)$.
From (ii), $\frac{dy}{dx} = 3x^2 - 24x + 20$. At $x=8$, gradient is $20$.
Equation of tangent: $y - (-96) = 20(x - 8) \implies y = 20x - 256$.
At $y=0$, $20x = 256 \implies x = 12.8$. So $C$ is $(12.8, 0)$.

2. Calculate the area of the shaded region above the $x$-axis (from $x=0$ to $x=2$)

$\text{Area}_1 = \int_0^2 (x^3 - 12x^2 + 20x) \, dx$
$= \left[ \frac{x^4}{4} - 4x^3 + 10x^2 \right]_0^2$
$= \left( \frac{16}{4} - 4(8) + 10(4) \right) - 0$
$= 4 - 32 + 40 = 12$

3. Calculate the area of the shaded region below the $x$-axis bounded by the curve (from $x=2$ to $x=8$)

$\text{Area}_2 = \left| \int_2^8 (x^3 - 12x^2 + 20x) \, dx \right|$
$= \left| \left[ \frac{x^4}{4} - 4x^3 + 10x^2 \right]_2^8 \right|$
$= \left| \left( \frac{4096}{4} - 4(512) + 10(64) \right) - 12 \right|$
$= |(1024 - 2048 + 640) - 12|$
$= |-384 - 12| = |-396| = 396$

4. Calculate the area of the shaded region bounded by the tangent line and the $x$-axis (from $x=8$ to $x=12.8$)

This region is a right-angled triangle with vertices at $(8, 0)$, $(12.8, 0)$, and $(8, -96)$.
$\text{Area}_3 = \frac{1}{2} \times \text{base} \times \text{height}$
$= \frac{1}{2} \times (12.8 - 8) \times 96$
$= \frac{1}{2} \times 4.8 \times 96 = 230.4$

5. Calculate the total shaded area

$\text{Total Area} = \text{Area}_1 + \text{Area}_2 + \text{Area}_3$
$= 12 + 396 + 230.4 = 638.4$