曲线方程为 $y = ax^3 + b$，其中 $a$ 和 $b$ 为常数。该曲线在 $x = 1$ 处的法线方程为 $5y + 2x = 12$。求 $a$ 和 $b$ 的值。

1. Find the gradient of the normal from its equation.

The equation of the normal is $5y + 2x = 12$.
Rearranging to the form $y = mx + c$:
$5y = -2x + 12$
$y = -\frac{2}{5}x + \frac{12}{5}$
The gradient of the normal is $-\frac{2}{5}$.

2. Find the gradient of the tangent at $x = 1$.

Let the gradient of the tangent be $m_t$.
$m_t \times \left(-\frac{2}{5}\right) = -1$
$m_t = \frac{5}{2}$

3. Differentiate the equation of the curve.

$y = ax^3 + b$
$\frac{dy}{dx} = 3ax^2$

4. Equate the derivative at $x = 1$ to the gradient of the tangent to find $a$.

At $x = 1$, $\frac{dy}{dx} = 3a(1)^2 = 3a$.
$3a = \frac{5}{2}$
$a = \frac{5}{6}$

5. Find the $y$-coordinate of the point on the curve at $x = 1$.

The normal passes through the point of contact on the curve. Substitute $x = 1$ into the normal equation:
$5y + 2(1) = 12$
$5y = 10$
$y = 2$
The point of contact is $(1, 2)$.

6. Substitute the point and $a$ into the curve equation to find $b$.

Substitute $x = 1$, $y = 2$, and $a = \frac{5}{6}$ into $y = ax^3 + b$:
$2 = \left(\frac{5}{6}\right)(1)^3 + b$
$b = 2 - \frac{5}{6} = \frac{7}{6}$