<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] Hint The product of two roots is $-1$ and the product of the other two roots is $1$.
Therefore
$$x^4-2x^3-3x^2+4x-1=(x^2+ax+1)(x^2+bx-1)$$
Oppening the brackets gives
$$a+b=-2\\
ab=-3 \\
b-a=4$$
which is trivial to solve.
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