[tex]f = $ x$^3-2$x$^2-2$x $+1 \\ \\ $x$_1+$x$_2+$x$_3 = -\dfrac{b}{a} \Rightarrow $ x$_1+$x$_2+$x$_3 = -\dfrac{-2}{1} \Rightarrow \\ \\ \Rightarrow
$ x$_1+$x$_2+$x$_3 = 2 $ $ $ $($scoatem 3 relatii din relatia asta:) \\ \\ $\boxed{1}\quad $x$_2+$x$_3 = 2-$x$_1 \\ \\ \boxed{2} \quad$x$_3+$x$_1 = 2-$x$_2\\ \\ \boxed{3} \quad $x$_1+$x$_2 = 2-$x$_3 \\ \\ \\ [/tex]
[tex]\Rightarrow ($x$_2+$x$_3)($x$_3+$x$_1)($x$_1+$x$_2) = (2-$x$_1)(2-$x$_2)(2-$x$_3) \\ \\ $Polinomul $ $x$^3-2$x$^2-2$x$+1 $ se mai poate scrie sub forma de \\ factori ireductibili ca: $($x$-$x$_1)($x$-$x$_2)($x$-$x$_3)[/tex]
[tex]f = $ x$^3-2$x$^2-2$x$+1 \\ f(2) = 2^3-2\cdot 2^2-2\cdot 2 + 1 = 8 -8-4+1 = -3 \\ \\ f = ($x$-$x$_1)($x$-$x$_2)($x$-$x$_3)\\ f(2) = (2-$x$_1)(2-$x$_2)(2-$x$_3) \\ \\ $Noi calcularam ca f(2) = -3 \Rightarrow (2-$x$_1)(2-$x$_2)(2-$x$_3) = -3[/tex]
[tex]\Rightarrow \boxed{($x$_2+$x$_3)($x$_3+$x$_1)($x$_1+$x$_2) =-3}[/tex]
