<s>[INST] Given $a+b=8$, $ax+by=9$, $ax^2+by^2=33$, $ax^3+by^3=60$, find $ax^4+by^4$ 
Given $$\begin{align}a+b&=8\\ax+by&=9\\ax^2+by^2&=33\\ax^3+by^3&=60\end{align}$$
Find $ax^4+by^4$

I really have no idea here.
Without coefficients $a$ and $b$, we can simply figure out $xy$ and find what we want by linear recurrences, but that doesn't seem to work here.
 [/INST] Following the hint in comments, letting $x+y=s$ and $xy=p$
we can write the following linear system:
$$60=33s-9p\\33=9s-8p$$
Solving gives $s=1,p=-3$ and
$$ax^4+by^4=60s-33p=159$$
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