### 2018 CMI Entrance Exam #2

$\textbf{(a)}$ Find all real solutions of the equation$$\Big(x^2-2x\Big)^{x^2+x-6}=1$$Explain why your solutions are the only solutions.

$\textbf{(b)}$ The following expression is a rational number. Find its value.$$\sqrt[3]{6\sqrt{3}+10} -\sqrt[3]{6\sqrt{3}-10}$$

a) $\Big(x^2-2x\Big)^{x^2+x-6}=1$ implies either  or $x^2-2x=1$. Now, there are two possible cases.

$\bullet~$ For the first case, we have $x^2+x-6=0$, as this would give $1$ when plugged into the final expression. So, $(x-2)(x+3)=0~\implies~x=2,-3$. But $x=2$ is an extraneous solution. So, $x=-3$ for this case.

$\bullet~$ For the second case, we need $x^2-2x=1$. Then, by the quadratic formula, $x=\frac{2\pm 2\sqrt{2}}{2}=1\pm \sqrt{2}$.

$\bullet~$ Lastly, we can have $x^2-2x=-1$ and $x^2+x-6$ is even. Solving, we see that $x=1$ and $1^2+1-6=-4$ which is even.

So, there are only $3$ distinct solutions, namely $$1~,~-3~,~\Big(1+\sqrt{2}\Big)~,~\Big(1- \sqrt{2}\Big).$$

b) Suppose that $a\sqrt3+b=\sqrt[3]{6\sqrt3+10}$ for integers $a$ and $b$. Cubing both sides, $$3\sqrt3a^3+9a^2b+3\sqrt3ab^2+b^3=6\sqrt3+10$$From this, we have $$\begin{cases}9a^2b+b^3=10\\a^3+ab^2=2\end{cases}$$It turned out that $(a,b)=(1,1)$ satisfy these equations, which means that $\sqrt[3]{6\sqrt3+10}=\sqrt3+1$ Similarly, $\sqrt[3]{6\sqrt3-10}=\sqrt3-1$. Therefore, $$\sqrt[3]{6\sqrt3+10}-\sqrt[3]{6\sqrt3-10}=(\sqrt3+1)-(\sqrt3-1)=\boxed2$$