A curve has the equation $y = x^3 + kx^2 + kx + 8$. Find the set of values of $k$ such that

1. Find the first derivative of the curve equation.

$\frac{dy}{dx} = 3x^2 + 2kx + k$

2. State the condition for the curve to be an increasing function.

For an increasing function, $\frac{dy}{dx} \geqslant 0$ for all real values of $x$.
$3x^2 + 2kx + k \geqslant 0$

3. Apply the discriminant condition for the quadratic to be always non-negative.

Since the coefficient of $x^2$ is $3 > 0$, the quadratic is non-negative for all $x$ if its discriminant is less than or equal to $0$.
$b^2 - 4ac \leqslant 0$
$(2k)^2 - 4(3)(k) \leqslant 0$
$4k^2 - 12k \leqslant 0$

4. Solve the quadratic inequality to find the range of values for $k$.

$4k(k - 3) \leqslant 0$
$0 \leqslant k \leqslant 3$

1. Find the derivative of the curve and set it to zero to find stationary points.

$\frac{dy}{dx} = 3x^2 + 2kx + k$
For stationary points, $\frac{dy}{dx} = 0$:
$3x^2 + 2kx + k = 0$

2. Apply the discriminant condition for exactly one stationary point.

For exactly 1 stationary point, the quadratic equation $3x^2 + 2kx + k = 0$ must have exactly one real root (repeated root).
Discriminant, $b^2 - 4ac = 0$
$(2k)^2 - 4(3)(k) = 0$
$4k^2 - 12k = 0$

3. Solve for $k$.

$4k(k - 3) = 0$
$k = 0 \text{ or } k = 3$