Solve the following equations: A) 4^(2x2+x−6) = 1, B) (x2−3x+1)6 = 1, C) (x2-3x+1)^(2x2+x−6) = 1

Equation A is $4^{(2x^2+x-6)} = 1$.
For an equation of the form $a^b = 1$, if $a \neq 1$ and $a \neq -1$, then the exponent $b$ must be 0.
Here, the base is 4, so we set the exponent to 0:

$$2x^2+x-6 = 0$$

We can solve this quadratic equation by factoring. We look for two numbers that multiply to $2 \times (-6) = -12$ and add to 1. These numbers are 4 and $-3$.

$$2x^2+4x-3x-6 = 0$$ $$2x(x+2)-3(x+2) = 0$$ $$(2x-3)(x+2) = 0$$

Setting each factor to zero:

$$2x-3 = 0 \implies 2x = 3 \implies x = \frac{3}{2}$$ $$x+2 = 0 \implies x = -2$$

a) Solutions for equation A:

$$x = \frac{3}{2}, -2$$

Equation B is $(x^2-3x+1)^6 = 1$.
For an equation of the form $B^E = 1$, where $E$ is an even integer (here $E=6$), the base $B$ can be 1 or $-1$.

Case 1: The base is 1.

$$x^2-3x+1 = 1$$ $$x^2-3x = 0$$ $$x(x-3) = 0$$

This gives $x=0$ or $x=3$.

Case 2: The base is $-1$.

$$x^2-3x+1 = -1$$ $$x^2-3x+2 = 0$$

Factoring the quadratic:

$$(x-1)(x-2) = 0$$

This gives $x=1$ or $x=2$.

b) Solutions for equation B:

$$x = 0, 1, 2, 3$$

Equation C is $(x^2-3x+1)^{(2x^2+x-6)} = 1$.
Let the base be $B = x^2-3x+1$ and the exponent be $E = 2x^2+x-6$.
We need to solve $B^E = 1$. There are three cases:

Case 1: The exponent is 0, provided the base is not 0.

$$E = 2x^2+x-6 = 0$$

From Step 1, the solutions are $x = \frac{3}{2}$ and $x = -2$.
We check the base $B$ for these values:
For $x = \frac{3}{2}$: $B = \left(\frac{3}{2}\right)^2 - 3\left(\frac{3}{2}\right) + 1 = \frac{9}{4} - \frac{9}{2} + 1 = \frac{9-18+4}{4} = -\frac{5}{4}$. Since $B \neq 0$, $x = \frac{3}{2}$ is a solution.
For $x = -2$: $B = (-2)^2 - 3(-2) + 1 = 4 + 6 + 1 = 11$. Since $B \neq 0$, $x = -2$ is a solution.

Case 2: The base is 1.

$$B = x^2-3x+1 = 1$$

From Step 2, the solutions are $x = 0$ and $x = 3$.
For these values, $1^E = 1$ holds for any real exponent $E$. So, $x=0$ and $x=3$ are solutions.

Case 3: The base is $-1$, and the exponent is an even integer.

$$B = x^2-3x+1 = -1$$

From Step 2, the solutions are $x = 1$ and $x = 2$.
We check if the exponent $E$ is even for these values:
For $x = 1$: $E = 2(1)^2 + (1) - 6 = 2 + 1 - 6 = -3$. Since $-3$ is an odd integer, $x=1$ is not a solution.
For $x = 2$: $E = 2(2)^2 + (2) - 6 = 8 + 2 - 6 = 4$. Since 4 is an even integer, $x=2$ is a solution.

Combining all valid solutions from the three cases:

$$x = -2, 0, \frac{3}{2}, 2, 3$$

c) Solutions for equation C:

$$x = -2, 0, \frac{3}{2}, 2, 3$$

a) Equation A:
From Step 1, there are 2 solutions: $x = \frac{3}{2}, -2$.
Number of values: $\boxed{\text{2}}$

b) Equation B:
From Step 2, there are 4 solutions: $x = 0, 1, 2, 3$.
Number of values: $\boxed{\text{4}}$

c) Equation C:
From Step 3, there are 5 solutions: $x = -2, 0, \frac{3}{2}, 2, 3$.
Number of values: $\boxed{\text{5}}$