If a,b,c are the roots of x^3-2x^2+x-3=0, find the value of

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  1. You might recall that for any quadratic of the form
    ax^2 + bx + c = 0 , if the roots are r1 and r2
    then r1+r2 = -b/a and r1r2 = c/a

    A similar relation exists for a cubic of the form
    ax^3 + bx^2 + cx + d = 0 , with roots r1, r2, and r3
    r1 + r2 + r3 = -b/a
    r1r2r3 = -d/a
    and also
    r1r2 + r1r3 + r2r3 = c/a

    so in your case
    a+b+c = 2
    abc = 3
    ab + ac + bc = 1

    then (b+c)(c+a)(a+b)
    = (b+c)(ac + bc + a^2 + ab)
    = abc + b^2c + a^2b + ab^2 + ac^2 + bc^2 + a^2c + abc
    = .....

    see if you can anywhere from here.

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