Solve for real $x$ : \[ (x^2 + x -2 )^3 + (2x^2 - x -1)^3 = 27(x^2 -1 )^3. \]
1. Given the equation:
   \[
   (x^2 + x - 2)^3 + (2x^2 - x - 1)^3 = 27(x^2 - 1)^3
   \]

2. We rewrite the equation in the form:
   \[
   (x^2 + x - 2)^3 + (2x^2 - x - 1)^3 + (3 - 3x^2)^3 = 0
   \]

3. Notice that the sum of the bases is zero:
   \[
   (x^2 + x - 2) + (2x^2 - x - 1) + (3 - 3x^2) = 0
   \]

4. Using the identity \(a^3 + b^3 + c^3 = 3abc\) when \(a + b + c = 0\), we can rewrite the equation as:
   \[
   3 (x^2 + x - 2) (2x^2 - x - 1) (3 - 3x^2) = 0
   \]

5. This implies that:
   \[
   (x^2 + x - 2) (2x^2 - x - 1) (3 - 3x^2) = 0
   \]

6. We solve each factor separately:
   - For \(x^2 + x - 2 = 0\):
     \[
     x^2 + x - 2 = (x - 1)(x + 2) = 0
     \]
     Thus, \(x = 1\) or \(x = -2\).

   - For \(2x^2 - x - 1 = 0\):
     \[
     2x^2 - x - 1 = (2x + 1)(x - 1) = 0
     \]
     Thus, \(x = -\frac{1}{2}\) or \(x = 1\).

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

7. Collecting all the solutions, we have:
   \[
   x = 1, x = -1, x = -2, x = -\frac{1}{2}
   \]

The final answer is \(\boxed{x = 1, x = -1, x = -2, x = -\frac{1}{2}}\).