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.
 A: 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.
A: 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?
A: Doing it in OP's way
$f(x)=x^4-2x^3-3x^2+4x-1=0$
let its rootsa be $a,b,c,d$, and let $a+b=u$ and $ab=v.$
Then by Vieta's formulas:
$$a+b+c+d=2~~~(1) \implies c+d=2-u$$
$$abcd=-1 ~~~~~(2) \implies cd=-1/v$$
$$ab+bc+cd+ac+bd+ad=-3~~~(3) \implies v-1/v+(a+b)(c+d)=-3 \implies v-1/v+u(2-u)=-3$$
$$abc+bcd+acd+bcd=-4~~~(4) \implies ab(c+d)+cd(a+b) =-4 \implies v(2-u)-(1/v)u=-4$$By putting $v=1$ in (3) we get $u^2-2u-3=0 \implies u=3,-1$
Next $a+b=3, ab=1; a+b=-1,ab=1$ give $$a,
b=\frac{3\pm \sqrt{5}}{2};~~ a,b=\frac{-1\pm \sqrt{5}}{2}$$
These for are the roots which can ve arranged ascending order as
$$x=\frac{-1-\sqrt{5}}{2},\frac{3-\sqrt{5}}{2},\frac{-1+\sqrt{5}}{2},\frac{3+\sqrt{5}}{2}~~~~(5)$$
Interestingly, (4) when $v=1$  also gives $u=3$, again. 
