<s>[INST] The equation $x^4-2x^3-3x^2+4x-1=0$ has four distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1The equation $x^4-2x^3-3x^2+4x-1=0$ has four distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1<x_2<x_3<x_4$ and product of two roots is unity, then:
$Q-1$: Find $x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4$
$Q-2$: Find $x_2^3+x_4^3$
My attempt is as follows:-
$A-1$ : First I tried to find any trivial root, but was not able to find any. After that I tried following:-
$$x_1\cdot x_2+x_1\cdot x_3+x_1\cdot x_4+x_2\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_1\cdot x_4-x_2\cdot x_3$$
$$x_1\cdot x_2\cdot x_3\cdot x_4=-1$$
$$x_1\cdot x_4=\dfrac{-1}{x_2\cdot x_3}$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_1\cdot x_4-x_2\cdot x_3$$
$$x_1\cdot x_2+x_1\cdot x_3+x_2\cdot x_4+x_3\cdot x_4=-3-x_2\cdot x_3+\dfrac{1}{x_2\cdot x_3}$$
But from here I was not able to proceed as I was not able to calculate value of $x_2\cdot x_3$
$A-2$ : $(x_2+x_4)(x_2^2+x_4^2-x_2\cdot x_4)$
Now here I was not getting any idea for how to proceed.
Please help me in this.
 [/INST] Also, we can use the following way.
For any value of $k$ we obtain: $$x^4-2x^3-3x^2+4x-1=(x^2-x+k)^2-x^2-k^2+2kx-2kx^2-3x^2+4x-1=$$
$$=(x^2-x+k)^2-((2k+4)x^2-(2k+4)x+k^2+1),$$ which for $k=0$ gives:
$$x^4-2x^3-3x^2+4x-1=(x^2-x)^2-(2x-1)^2=(x^2-3x+1)(x^2+x-1).$$
Can you end it now?
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