<s>[INST] Determine $ax^4 + by^4$ for system of equations I found the following recreational problem without further specification for $a,b$. 
Let $x,y$ be real numbers s.t.
$a + b = 6$, 
$ax + by = 10$, 
$ax^2 + by^2 = 24$,
$ax^3 + by^3 = 62$.
Determine $ax^4 + by^4$. 
I am new to problem solving exercises like this and therefore appreciate diverse approaches to this problem as well as comments on how to tackle those types of exercises. 
 [/INST] $$ \color{red}{3 \cdot 62 - 24 = 162}.   $$
$$ 3 \cdot 10 - 6 = 24.   $$
$$ 3 \cdot 24 - 10 = 62.   $$
This is the observation of @Ross,
$$  (xy-1)(x-y)^2 = 0.  $$ Ross points out in comment (below) that an ingredient was $44=abxy(x-y)^2,$ so that
$$ a \neq 0, b \neq 0, x \neq 0, y \neq 0, x \neq y. $$ Finally,
$$  xy = 1.  $$
The simple fact I got was
$$ a (x^2 - 3x+1) + b ( y^2 - 3 y + 1) = 0.  $$
We have $xy = 1.$ If $x^2 - 3 x + 1 \neq 0$ and $y^2 - 3 y + 1 \neq 0,$
$$  y^2 - 3 y + 1 = (x^2 - 3 x + 1) / x^2,$$
$$ x^2 ( y^2 - 3 y + 1 ) = x^2 - 3 x + 1.  $$
$$ a x^2 (y^2 - 3 y + 1) + b (y^2 - 3 y + 1) = 0,$$
$$ (a x^2 + b)(y^2 - 3 y + 1) = 0.  $$
Switch the letters, we get
$$ (a  + b y^2)(x^2 - 3 x + 1) = 0.  $$
Alright,
$$ a x^2 + b y^2 = 24, a x^2 + b = 0.  $$
$$  b(y^2 - 1) = 24. $$ Also
$$  a ( x^2 - 1) = 24. $$
Add,
$$   a(x^2 - 1) + b ( y^2 - 1) = 48.$$ However, we know
$$ a x^2 + b y^2 - a - b = 24 - 6 = 18.  $$ This contradicts the assumption 
$ x^2 - 3 x + 1 \neq 0.   $
$$  \color{red}{ x^2 - 3 x + 1 = 0} $$
$$ \color{red}{  y^2 - 3 y + 1 = 0} $$
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