<s>[INST] Finding $\cot(a+b)$, if $\tan^3a+2\tan^3b=6$, $\tan^4a+2\tan^4b=18$, $\tan^5a+2\tan^5b=30$, and $\tan^6a+2\tan^6b=66$ A "Brain Teaser" with Trigonometry:

If
$$\begin{align}
\tan^3 (\alpha) +2\tan^3 (\beta)&=\phantom{0}6 \\
\tan^4 (\alpha) +2\tan^4 (\beta)&=18 \\
\tan^5 (\alpha) +2\tan^5 (\beta)&=30 \\
\tan^6 (\alpha) +2\tan^6 (\beta)&=66
\end{align}$$
Then find $\cot(\alpha+\beta)$.

Answer is $3$.
I have tried to form terms of $\tan(\alpha+\beta)$ by multiplying 1st term with $\tan(\alpha) +\tan(\beta)$ and again 1st term with $\tan^2(\alpha) +\tan^2(\beta)$ to get relationship with different terms hoping to extract $\tan(\alpha) \tan(\beta)$ and cancel out terms but no use.
Any suggestions???
 [/INST] Something of a trick here. If you render $6=8-2,18=16+2,30=32-2,66=64+2$ you discover that the equations are satisfied by rendering $\tan\alpha=2,\tan\beta=-1$.
Then apply the identity
$\tan(\alpha+\beta)=\dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
and proceed from there.

There is also a way to extract $\tan\alpha$ and $\tan\beta$ without "inspection".  Let these parameters be $a$ and $b$ respectively and consider first the two equations
$a^3+2b^3=6$
$a^6+2b^6=66$
From the first equation render $a^3=6-2b^3$, thus substitute $(6-2b^3)^2$ for $a^6$ into the second equation. This leads to
$b^6-4b^3-5=0$
which can be solved as a quadratic equation for $b^3$. Thus $b^3\in\{-1,5\}$ and $a^3=6-2b^3$, from which we then obtain the two candidate solutions
$a=\tan\alpha=2, b=\tan\beta=-1$
$a=\tan\alpha=-\sqrt[3]4, b=\tan\beta=\sqrt[3]5$
The second candidate, with its cube-root surds, cannot give the required whole number sums for $a^4+2b^4$ and $a^5+2b^5$, so only the first one checks with all four equations. We then proceed with the tangent formula as above.
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