<s>[INST] Finding the value of $ax^4+by^4$ 
If $\quad a+b=23 , \quad ax+by=79,\quad ax^2+by^2=217,\quad
 ax^3+by^3=691\quad$ find the value of $ax^4+by^4$.

Here is my attempt:
$$(a+b)(x+y)=ax+by+ay+bx\rightarrow 23(x+y)=79+(ay+bx)$$
$$(ax+by)(x+y)=ax^2+by^2+axy+bxy\rightarrow79(x+y)=217+23 xy$$
In each equation I have two unknowns it seems that doesn't work.
 [/INST] Answer to OP's comment:
For solving the following system:

$$79(x+y)=23xy+217$$$$ 217(x+y)=79xy+691$$

We first eliminate $(x+y) $ from both equations:
$$ (x+y) = \frac{1}{79} (23 xy + 217)\tag{1}$$
And,
$$ x+y = \frac{1}{217} (79 xy + 691) \tag{2}$$
We get:
$$  \frac{1}{217} (79 xy + 691)= \frac{1}{79} (23 xy + 217)$$
We get the result
$$ y = \frac{-6}{x} \tag{3}$$
Now we can substitute (3) into (1), and then if we multiply the whole equation by $x$ , we can solve for the quadratic in x
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