If a,b,c,are three roots of xpower3+pxpower2+qx+r,find sum of(a square b+b square c+ c square a)?

if a,b,c are the roots of xpower3+pxpower2+qx+r find som of(a square *b+b square*c+c square *a) and sum of (a power3+b power3+c power3)

If a,b,c,are three roots of xpower3+pxpower2+qx+r,find sum of(a square b+b square c+ c square a)?
a^2 b +b^2 c+c^2 a cannot be written in terms of p,q, and r because it is not a symmetric function of a,b, and c. In particular, interchanging a and b makes the expression different.





However, a^2 b+a^2 c+ b^2 a +b^2 c +c^2 a +c^2 b


=(a+b+c)(ab+ac+bc)-3abc


=-pq+3r.
Reply:the third degree equation x^3+px^2+qx+r has three roots a,b,c


a+b+c=-p


ac+ab+cb=q


abc=-r


a^3+b^3+c^3=(a+b+c)^3-3*


(a^2*b+b^2*a+c^2*b+b^2*c+


a^2*c+c^2*a)-


6(abc)


=(a+b+c)^3-3*[((a+b+c)^2


-2(ab+cb+ac))*


(a+b+c))-3*abc)]-3*abc


=(-p)^3-3*[((-p)^2-2*(q))*(-p))


+3*r]+6*r





I don't think that you can obtain


a^*b+b^2*c+c^2*a


would you give me 10 points?
Reply:a^3+b^3+c^3=


(a+b+c)^3 - 3(a^2*b+b^2*a+c^2*b+b^2*c+a^2*c+c^2*a) - 3(abc)





I will denote (a^2*b+b^2*a+c^2*b+b^2*c+a^2*c+c^2*a) by x so that I dont have to rewrite it all the time





Also from the cubic equation,





a+b+c = -p


ac + ab + bc = q


abc = -r





Then,


-pq = x + 3abc





x = 3r - pq





so, a^3 + b^3 + c^3 = -p^3 + 9r - 3pq + 3r


= -p^3 - 3pq + 12r
Reply:U MUST HAVE BEEN REALLY BOOOOOORED