Calculate the value of the expression for polynomial roots The roots of the polynomial equation: $ x ^ 3 - 3x -1 = 0 $ are: $ a, b $ and $ c $. Based on this, calculate the value of:$$ L = \frac {1} {a ^ 2-2a + 2} + \frac {1} {b ^ 2-2b + 2} + \frac {1} {c ^ 2-2c + 2} $$
Attemp:
$L=\frac{1}{(a-1)^2+1}+\frac{1}{(b-1)^2+1}+\frac{1}{(c-1)^2+1}$.
First, we want to find a polynomial whose roots are $a-1$, $b-1$ and $c-1$. This polynomial is just $(x+1)^3-3(x+1)-1=x^3+3x^2-3$. If we say the roots of this are $a'$, $b'$ and $c'$, we want to find $\frac{1}{a'^2+1}+\frac{1}{b'^2+1}+\frac{1}{c'^2+1}$.
Now we want to find what polynomial has roots $a'^2$, $b'^2$ and $c'^2$. In order to find what this polynomial is, we need to know $a'^2+b'^2+c'^2$, $a'^2b'^2+b'^2c'^2+c'^2a'^2$ and $a'^2b'^2c'^2$. $a'^2+b'^2+c'^2=(a'+b'+c')^2-2(a'b'+b'c'+c'a')=3^2-2(0)=9$. $a'^2b'^2+b'^2c'^2+c'^2a'^2=\frac{(a'^2+b'^2+c'^2)^2-a'^4-b'^4-c'^4}{2}$. To find $a'^4+b'^4+c'^4$, we can realize that $a'^3+b'^3+c'^3=(a'^2+b'^2+c'^2)(a'+b'+c')-(a'+b'+c')(a'b'+b'c'+c'a')+$ $a'b'c'(a'^0+b'^0+c'^0)=-3(a'^2+b'^2+c'^2)+3(a^0+b^0+c^0)=-18$, and similarly $a'^4+b'^4+c'^4=-3(-18)+3(-3)=45$. Our expression is therefore $\frac{(a'^2+b'^2+c'^2)^2-a'^4-b'^4-c'^4}{2}=\frac{81-45}{2}=18$. Also, it's easy to see $(a'b'c')^2=9$. Our polynomial is $x^3-9x^2+18x-9$
Our sum is the sum of the reciprocals of the $(x-1)^3-9(x-1)^2+18(x-1)-9=x^3-12x^2+39x-37$ which equals $\frac{39}{37}$
attemp2: I also think about using the logarithmic derivative you can think that
$\frac{1}{(1 - a) ^2} + \frac{1}{(1 - b)^2} + \frac{1 }{(1 - c)^2} = \frac{(p (1) p '' (1) - p '(1)^2) } {p (1 )^2} = 2$, then there must be some magical way to make that 1 appear by adding to the denominator, without having to appeal to polynomial transforms
Am I right? Is there an easier way to solve this problem?

This page gives a nice summary of common polynomial root transforms, which we can employ here.
First, let's find a polynomial that has roots that are $a-1, b-1, c-1$. This is just $$Q(x) = P(x+1) = (x+1)^3 -3(x+1) -1 = x^3 +3x^2 -3.$$
Next, we find the polynomial that has roots $(a-1)^2, (b-1)^2, (c-1)^2.$ This is
$$R(x) = -Q(\sqrt{x}) Q(-\sqrt{x}) = x^3 - 9x^2 + 18x - 9$$
Then, the polynomial having roots $(a-1)^2 + 1, (b-1)^2 + 1, (c-1)^2 +1$ is
$$S(x) = R(x-1) = x^3 - 12x^2 + 39x - 37.$$
We could find the polynomial that has roots that are the reciprocals of the roots of $S(x)$, however, we can apply Vieta to obtain that the sum is indeed $\frac{39}{17}.$
What you did is of course correct, but you were more explicit with these computations. These tricks are nice to know so you don't have to do all the work every time!
As for your second approach, it's not clear to me how to proceed. Maybe someone else could shed some light?
