theory of equation

Question

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

Answer 1

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.

Answer 2

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?

Answer 3

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: $$x^4-(a+b+c+d)x^3+(cd+bc+bd+ad+ac+ab)x^2-(abc+abd+acd+bcd)x+abcd$$

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

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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 :)