parmenides51 wrote: Let $f(x) = x^3 + 3x^2 - 1$ have roots $a,b,c$. (a) Find the value of $a^3 + b^3 + c^3$ (b) Find all possible values of $a^2b + b^2c + c^2a$ The three roots of $P(x)=x^3-2x^2-x+1$ are $a>b>c \in R$. Find the value of $a^2b+b^2c+c^2a$

By Vieta's Formula, a+b+c=-3; ab+bc+ca=0; abc=1 (i) Σa³-3abc=(Σa)(Σa²-Σab)=> Σa³=(-3)(9)+3=-24 (ii)|(a²b+b²c+c²a)-(ab²+bc²+ca²)|=|(a-b)(b-c)(c-a)| Let's take, S1=a²b+b²c+c²a & S2=ab²+bc²+ca² Now we have to find an EQUATION with roots S1 & S2. (Sum of roots)=Σa²(b+c)=> Σa²(-3-a)=-[3(Σa²)+(Σa³)]=-3 (Product of roots)=Σa³b³+3(abc)²+Σabc(Σa³)=3+3-24=-18 Therefore, The Quadratic is x²+3x-18=0 whose roots are -6 & 3 Therefore, a²b+b²c+c²a=-6 or 3 Nice one.