TITLE: Finding $\cot(a+b)$, if $\tan^3a+2\tan^3b=6$, $\tan^4a+2\tan^4b=18$, $\tan^5a+2\tan^5b=30$, and $\tan^6a+2\tan^6b=66$
QUESTION [6 upvotes]: 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???

REPLY [4 votes]: There is no need to find $\tan (\alpha)$ and $\tan(\beta)$ separately, only their sum and product, because
$$
\cot(\alpha+\beta)
=\dfrac{1-\tan(\alpha)\tan(\beta)}{\tan(\alpha)+\tan(\beta)}
$$
Let $u_n = \tan^n (\alpha) +2\tan^n (\beta)$. Then $u_{n+2}=s u_{n+1}-p u_n$, where $s=\tan (\alpha)+\tan(\beta)$ and $p=\tan (\alpha)\tan(\beta)$, because $\tan(\alpha)$ and $\tan(\beta)$ are the roots of $x^2=sx-p$.
Then $$u_6 = su_5-pu_4, \quad u_5=su_4-pu_3$$ gives $$66 = 30s-18p, \quad 30=18s-6p$$ and so $p = -2$ and $s = 1$.
Therefore
$$
\cot(\alpha+\beta)
=\dfrac{1-\tan(\alpha)\tan(\beta)}{\tan(\alpha)+\tan(\beta)}
=\frac{1-(-2)}{1}=3
$$