Solve for x: $(x^2 - 3x + 1)^{(2x^2 + x - 6)} = 1$

Steps to Solve

1. Case 1: The exponent is zero If the exponent $2x^2 + x - 6 = 0$ and the base $x^2 - 3x + 1 \neq 0$, then the equation becomes $1$.

2. Solve the quadratic exponent equation Solve $2x^2 + x - 6 = 0$: $$2x^2 + x - 6 = 0$$ $$(2x - 3)(x + 2) = 0$$ $$x = \frac{3}{2}, -2$$

3. Check the base for $x = \frac{3}{2}$ For $x = \frac{3}{2}$: $$(\frac{3}{2})^2 - 3(\frac{3}{2}) + 1 = \frac{9}{4} - \frac{9}{2} + 1 = \frac{9 - 18 + 4}{4} = \frac{-5}{4} \neq 0$$ Hence $x = \frac{3}{2}$ is a solution.

4. Check the base for $x = -2$ For $x = -2$: $$(-2)^2 - 3(-2) + 1 = 4 + 6 + 1 = 11 \neq 0$$ Hence $x = -2$ is a solution.

5. Case 2: The base is one If the base $x^2 - 3x + 1 = 1$, then the equation becomes $1$ regardless of the exponent.

6. Solve the quadratic base equation Solve $x^2 - 3x + 1 = 1$: $$x^2 - 3x = 0$$ $$x(x - 3) = 0$$ $$x = 0, 3$$

7. Case 3: The base is negative one and the exponent is even If the base $x^2 - 3x + 1 = -1$ and the exponent $2x^2 + x - 6$ is an even integer, then the equation becomes $1$.

8. Solve the quadratic base equation Solve $x^2 - 3x + 1 = -1$: $$x^2 - 3x + 2 = 0$$ $$(x - 1)(x - 2) = 0$$ $$x = 1, 2$$

9. Check the exponent for $x = 1$ For $x = 1$: $$2(1)^2 + 1 - 6 = 2 + 1 - 6 = -3$$ $-3$ is odd, so $x = 1$ is not a solution.

10. Check the exponent for $x = 2$ For $x = 2$: $$2(2)^2 + 2 - 6 = 8 + 2 - 6 = 4$$ $4$ is even, so $x = 2$ is a solution.

11. Gather all solutions The solutions are $x = -2, 0, \frac{3}{2}, 2, 3$.