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Find the sum of the squares of the three solutions of the equation x3+3x2−7x+1=0.

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    Assume the equation has three roots a,b and c:

    then
    (x-a)(x-b)(x-c)=0
    x³-(a+b+c)x²+(ab+bc+ca)x-abc=0
    which means
    (a+b+c)=-3 (negative of coeff.of x²)
    (ab+bc+ca)=-7 (coeff. of x)

    Hence
    a²+b²+c²
    =(a+b+c)²-2(ab+bc+ca)
    =3²-2(-7)
    =9-(-14)
    =23

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