What is the sum of all $x$ such that $(3x^2 + 9x - 2012)^{(x^3-2012x^2-10x+1)} = 1$?
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2Something both the answers (so far) have ignored: You also have a solution if the bottom is equal to 1, or if it is equal to -1 and the top is an even integer. – Aaron Apr 06 '13 at 22:11
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Note that $x^3-2012x^2-10x+1$ has no rational root (it would have to be $\pm1$, which can be checked explicitly). Also note that yb polynomial division $$\begin{align}x^3-2012x^2-10x+1 &= (3x^2+9x-2012)\cdot\frac{x-2015}3 + \frac{20117 x -4054177}3, \\ x^3-2012x^2-10x+k &= (3x^2+9x-2013)\cdot\frac{x-2015}3 + 6706 x -1352065+k, \\ x^3-2012x^2-10x+1 &= (3x^2+9x-2011)\cdot\frac{x-2015}3 + \frac{20116 x -4052162}3. \end{align}$$ The first implies that $3x^2+9x-2012$ and $x^3-2012x^2-10x+1$ cannot both be zero as that would lead to a rational root. Similarly, the third implies that we cannot have base $=1$ and exponent $=0$ at the same time. Finally, the second equation shows that an $x$ for which the base is $-1$ and the exponent an integer $1-k$, again $x$ must be rational. However, $3x^2+9x-2013$ has no rational solutions.
Therefore, no weird special cases occur and the desired sum is simply the sum of the distinct(!) roots of $x^3-2012x^2-10x+1$ and the distinct(!) roots of $3x^2+9x-2013$. These can be read directly from the coefficients so that we obtain $$2012-\frac 93=2009.$$
Hagen von Eitzen
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Recall that $a^0 = 1$ for all real $a \neq 0$, and $1^b = 1$ for all real $b$.
$$(3x^2 + 9x - 2012)^{\large(x^3-2012x^2-10x+1)} = 1 \iff$$
$$ $$
$$x^3 \color{blue}{\bf - 2012}x^2 - 10 x + 1 = 0\tag{1}$$
or $$3x^2 + 9x - 2012 = 1 \iff 3x^2 + 9x - 2013 = 0 \iff \color{red}{\bf 1}\cdot x^2 + \color{red}{\bf 3}x - 671 = 0\quad\quad\quad\quad\tag{2}$$
There exist $3$ roots $x_1, x_2, x_3$ to $(1)$, and $2$ roots $x_4, x_5$ to (2).
The sum $S$ you want is $$S = \color{blue}{\bf x_1 + x_2 + x_3} + \color{red}{\bf x_4 + x_5}$$
To obtain this, note the colored coefficients: $${\bf Sum} = \color{blue}{\bf 2012} - \color{red}{\bf \frac 31} = {\bf 2009}.$$
amWhy
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@amWhy: Very nice touch with the color coding to highlight important points! +1! – Amzoti Apr 07 '13 at 03:46
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By the relation between coefficients of polynomial $x^3-2012x^2-10x+1$ and its roots $\lambda_i$ we know that $$\lambda_1+\lambda_2+\lambda_3=2012$$
By the same method we have for the polynomial $3x^2 + 9x - 2012-1$ $$\lambda'_1+\lambda'_2=-3$$ so the desired sum is $$\lambda_1+\lambda_2+\lambda_3+\lambda'_1+\lambda'_2=2009$$