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The equation

{(x^{2} + x - 1)}^{(x^{2} - 7 x + 12)} = 1

has $n$ distinct solutions.The sum of these solutions is $S$ .

Give the value of

S^{n}

Hint

Possibilities include:

Solution

Case 1

In the first case, we consider $x^{2} + x - 1 = 1$ . Solving this gives

\begin{aligned} x^{2} + x - 1 & = 1 \\ x^{2} + x - 2 & = 0 \\ (x + 2) (x - 1) & = 0 \\ x & \in {- 2, 1} \end{aligned}

Case 2

In the second case, we consider $x^{2} + x - 1 \neq 0$ and $x^{2} - 7 x + 12 = 0$ .

\begin{aligned} x^{2} - 7 x + 12 & = 0 \\ (x - 4) (x - 3) & = 0 \\ x & \in {3, 4} \end{aligned}

We note that $x^{2} + x - 1 \neq 0$ when $x \in {3, 4}$ so both of these solutions are valid.

Case 3

Finaly, we consider $x^{2} + x - 1 = - 1$ and $x^{2} - 7 x + 12$ is even.

\begin{aligned} x^{2} + x - 1 & = - 1 \\ x^{2} + x & = 0 \\ x (x + 1) & = 0 \\ x & \in {- 1, 0} \end{aligned}

It turns out that $x^{2} - 7 x + 12$ is in fact even for $x \in {- 1, 0}$ and so both solutions are valid.

The total solution set is

x \in {- 2, 1, 3, 4, - 1, 0}