To solve the equations provided, we can break them down step by step.

### c) \((x^2 + x)^2 + 4(x^2 + x) - 12 = 0\)

1. Let \(y = x^2 + x\). Then the equation becomes:
\[
y^2 + 4y - 12 = 0
\]
2. This is a quadratic equation in \(y\). We can solve it using the quadratic formula:
\[
y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} = \frac{{-4 \pm \sqrt}}{2} = \frac{{-4 \pm \sqrt{64}}}{2} = \frac{{-4 \pm 8}}{2}
\]
- \(y = 2\) or \(y = -6\).

3. Substitute back \(x^2 + x = y\):
- For \(y = 2\):
\[
x^2 + x - 2 = 0 \implies (x - 1)(x + 2) = 0 \implies x = 1 \text{ or } x = -2
\]

- For \(y = -6\):
\[
x^2 + x + 6 = 0
\]
The discriminant is \(1 - 24 < 0\), meaning there are no real solutions.

### Solutions for c):
- \(x = 1\) and \(x = -2\).

---

### d) \(x(x - 1)(x^2 - x + 1) - 6 = 0\)

1. First, set \(x(x - 1)(x^2 - x + 1) = 6\). Look for values of \(x\).

2. Testing for integer values:
- For \(x = 0\):
\[
0(0-1)(0^2-0+1) = 0 \quad (\text{not a solution})
\]
- For \(x = 1\):
\[
1(1 - 1)(1^2 - 1 + 1) = 0 \quad (\text{not a solution})
\]
- For \(x = 2\):
\[
2(2 - 1)(2^2 - 2 + 1) = 2(1)(3) = 6 \quad (\text{solution: } x = 2)
\]

- For \(x = 3\):
\[
3(3 - 1)(3^2 - 3 + 1) = 3(2)(7) = 42 \quad (\text{too high})
\]

- For \(x = -1\):
\[
-1(-1 - 1)((-1)^2 + 1 + 1) = -1(-2)(3) = 6 \quad (\text{solution: } x = -1)
\]

### Solutions for d):
- \(x = 2\) and \(x = -1\).

In summary:
- **For equation c:** \(x = 1, -2\)
- **For equation d:** \(x = 2, -1\)