theory of equation let the roots of the equation $$x^4 -3x^3 +4x^2 -2x +1=0$$
be a , b, c,d
then find the value of
$$ (a+b) ^{-1} + (a+c) ^{-1}+ (a+d)^{-1} + (b+c)^{-1} +
 ( c+d)^{-1}+ (c+d)^{-1}$$
my solution i observed that the roots are imaginary roots of the
equation $$ (x-1)^5 =1$$. but after that i am stuck
 A: Notice that
$$\frac{1}{a+b} + \frac{1}{c+d} = \frac{a+b+c+d}{(a+b)(c+d)} = \frac{3}{(a+b)(c+d)}.$$
Repeat for two more sets of two fractions. Then add the three results.
A: If you combine your result with that of B. Goddard, you have
$$\frac{1}{a+b}+\frac{1}{c+d} = \frac{3}{(a+b)(c+d)}$$
Now suppose $a=1+w, b=1+w^2, c=1+w^3, d=1+w^4$ where $w$ is one of the imaginary roots of $w^5=1$, then
$$(a+b)(c+d) = (2+w+w^2)(2+w^3+w^4) \\= 4 + 2(w+w^2+w^3+w^4) + (w^4 + 1 + 1 + w)=4+w+w^4$$
Similarly,
$$(b+c)(a+d)=1$$
$$(a+c)(b+d)=4+w^2+w^3$$
That's probably easier since both $w+w^4$ and $w^2+w^3$ are real. You will need $\cos \left(\frac{2\pi}{5}\right)=\frac{\sqrt 5 -1}{4}$
Edit: Actually you don't need the cosine. Just add $1/(4+w+w^4)$ and $1/(4+w^2+w^3)$ like the above it becomes very clean.
$$(4+w^2+w^3)(4+w+w^4)\\=16+4(w+w^2+w^3+w^4)+(w^3+w+w^4+w^2)\\=16-4-1=11$$
Can you end it now?
A: If the polynomial has roots $a,b,c,d$ then we can write it as:
$$(x-a)(x-b)(x-c)(x-d)=0$$
which if we expand out we get:
$$x^4-(a+b+c+d)x^3+(cd+bc+bd+ad+ac+ab)x^2-(abc+abd+acd+bcd)x+abcd$$

From this we can take values from our polynomial and so:
$$a+b+c+d=3$$
$$cd+bc+bd+ad+ac+ab=4$$
$$abc+abd+acd+bcd=2$$
$$abcd=1$$

Now try and get a single expression from the desired product and you should be able to represent it in terms of the expressions above, giving you a value :)
