<s>[INST] Linearly express a polynomial of a root Consider the equation $x^2-4x+7=0$ and $x_1,x_2$ its roots. Calculate:
$$\frac{x_1^4-2x_1^3+3}{x_1^3+x_1^2}+\frac{x_2^4-2x_2^3+3}{x_2^3+x_2^2}$$
Rules: cannot manually calculate the roots, must use Vieta's formulas.
Amplifying the fractions in this form seems like a bad idea so I thought of a trick to simplify it.
Substitute $x_1$ in the main equation:
$$x_1^2-4x_1+7=0$$
Multiply by $x_1^2$ then add $2x_1^3+3$ and get: $x_1^4-2x_1^3+3=2x_1^3-7x_1^2+3$
For the same equation multiply by $2x_1$ then add $x_1^2+3$: $2x_1^3-7x_1^2+3=x_1^2-14x_1+3$
Now substract $10x_1+4$ so $x_1^2-14x_1+3=-(10x_1+4)$
$\Rightarrow x_1^4-2x_1^3+3=-(10x_1+4)$
Do you notice how in the chain of equations by each iteration we lose 1 degree?
Same thing for $x_1^3+x_1^2=13x_1-35$. Now the expression is easily computable.
My question is - is there an easier way to linearly express a polynomial of a root?
 [/INST] from $x^2 = 4x-7$  I got $x^3 = 9x - 28$ and $x^4 = 8x-63$
Next, $x = 4 - \frac{7}{x},$ so $\frac{1}{x}  = \frac{-x+4}{7}$ and  so $\frac{1}{x^2}  = \frac{-4x+9}{49}$
Then I wrote $(x+1)^2 = 6(x+1)  - 12; \; \;$  then dividing gives $x+1 = 6 - \frac{12}{x+1}$  and
$ \frac{1}{x+1}  = \frac{-x + 5}{12}  $
</s>