[剑桥/9709/2010] 图中显示了曲线 $y = 9 - x^3$ 和 $y = \frac{8}{x^3}$ 的部分图像以及它们的交点 $P$ 和 $Q$。$P$ 和 $Q$ 的 $x$ 坐标分别为 $a$ 和 $b$。

1. Set up the equation to find the points of intersection.

At the points of intersection, the $y$-values of both curves are equal:
$9 - x^3 = \frac{8}{x^3}$

2. Rearrange the equation to show the required form.

Multiplying both sides by $x^3$ (since $x \neq 0$ in the first quadrant):
$9x^3 - x^6 = 8$
Rearranging the terms to one side gives:
$x^6 - 9x^3 + 8 = 0$
Since the intersection points $P$ and $Q$ have $x$-coordinates $a$ and $b$, $x = a$ and $x = b$ must satisfy this equation and are therefore its roots.

3. Solve the polynomial equation by treating it as a quadratic in $x^3$.

Let $u = x^3$. The equation becomes a quadratic in $u$:
$u^2 - 9u + 8 = 0$
$(u - 1)(u - 8) = 0$
$u = 1 \quad \text{or} \quad u = 8$

4. Solve for $x$ and assign the values to $a$ and $b$.

Substitute back $u = x^3$:
$x^3 = 1 \implies x = 1$
$x^3 = 8 \implies x = 2$
From the given diagram, $P$ is to the left of $Q$, meaning $a < b$.
Therefore, $a = 1$ and $b = 2$.

1. Solve for the intersection points from part (i) to find the limits of integration.

Let $u = x^3$. The intersection equation $x^6 - 9x^3 + 8 = 0$ becomes $u^2 - 9u + 8 = 0$.
$(u - 1)(u - 8) = 0 \implies u = 1 \text{ or } u = 8$.
$x^3 = 1 \implies x = 1$
$x^3 = 8 \implies x = 2$
Since $P$ is to the left of $Q$, $a = 1$ and $b = 2$.

2. Set up the definite integral for the area between the two curves.

From the graph, the curve $y = 9 - x^3$ is above the curve $y = \frac{8}{x^3}$ for $1 \le x \le 2$.
$\text{Area} = \int_1^2 \left( 9 - x^3 - \frac{8}{x^3} \right) \, dx$

3. Integrate the function with respect to $x$.

$\int \left( 9 - x^3 - 8x^{-3} \right) \, dx = \left[ 9x - \frac{x^4}{4} - \frac{8x^{-2}}{-2} \right]_1^2$
$= \left[ 9x - \frac{x^4}{4} + \frac{4}{x^2} \right]_1^2$

4. Evaluate the definite integral to find the exact area.

Substitute $x = 2$ and $x = 1$ into the antiderivative:
$\text{Area} = \left( 9(2) - \frac{2^4}{4} + \frac{4}{2^2} \right) - \left( 9(1) - \frac{1^4}{4} + \frac{4}{1^2} \right)$
$= (18 - 4 + 1) - (9 - 0.25 + 4)$
$= 15 - 12.75$
$= 2.25$

1. Find the gradient functions for both curves by differentiating with respect to $x$.

Let $y_1 = 9 - x^3$ and $y_2 = \frac{8}{x^3} = 8x^{-3}$.
$\frac{dy_1}{dx} = -3x^2$
$\frac{dy_2}{dx} = -24x^{-4} = -\frac{24}{x^4}$

2. Set up the equation for parallel tangents at $x = c$.

Since the tangents to both curves at $x = c$ are parallel, their gradients at this point must be equal. Substituting $x = c$ into both derivatives and equating them:
$-3c^2 = -\frac{24}{c^4}$

3. Solve for $c$, considering the interval $a < c < b$.

$c^2 \cdot c^4 = \frac{-24}{-3}$
$c^6 = 8$
$c = \pm \sqrt[6]{8} = \pm 2^{\frac{3}{6}} = \pm \sqrt{2}$
From part (i), the roots of the intersection are $a = 1$ and $b = 2$.
Since $a < c < b$, we require $1 < c < 2$.
Therefore, the only valid solution is:
$c = \sqrt{2}$